Separable differential equations are first-order differential equations where the variables can be separated onto opposite sides of the equation. The goal is to rewrite the equation so that all \(y\) terms are with \(dy\) and all \(x\) terms are with \(dx\). After separating the variables, integrate both sides and solve for \(y\) when possible.
Practice Problems
\(\textbf{1)}\) \(\frac{dy}{dx}=-\frac{x}{y}\)
\(\,\,\,\,\, \frac{dy}{dx}=-\frac{x}{y} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, y dy=-x dx \,\,\,\left(\text{Multiply both sides by } y\, dx\right)\)
\(\,\,\,\,\, \int y dy= \int -x dx \,\,\,\left(\text{Take integral of both sides}\right)\)
\(\,\,\,\,\, \frac{y^2}{2} = \frac{-x^2}{2}+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y^2 = -x^2+2c \,\,\,\left(\text{Multiply both sides by }2 \right)\)
\(\,\,\,\,\, y^2 = -x^2+c \,\,\,\left(\text{Let }c=2c \right)\)
\(\,\,\,\,\, \sqrt{y^2} = \sqrt{-x^2+c} \,\,\,\left(\text{Square root both sides } \right)\)
\(\,\,\,\,\, y = \pm\sqrt{-x^2+c} \,\,\,\left(\text{Simplify} \right)\)
The answer is \( y = \pm\sqrt{-x^2+c} \)
\(\,\,\,\,\, \frac{dy}{dx}=-\frac{x}{y} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, y dy=-x dx \,\,\,\left(\text{Multiply both sides by } y\, dx\right)\)
\(\,\,\,\,\, \int y dy= \int -x dx \,\,\,\left(\text{Take integral of both sides}\right)\)
\(\,\,\,\,\, \frac{y^2}{2} = \frac{-x^2}{2}+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y^2 = -x^2+2c \,\,\,\left(\text{Multiply both sides by }2 \right)\)
\(\,\,\,\,\, y^2 = -x^2+c \,\,\,\left(\text{Let }c=2c \right)\)
\(\,\,\,\,\, \sqrt{y^2} = \sqrt{-x^2+c} \,\,\,\left(\text{Square root both sides } \right)\)
\(\,\,\,\,\, y = \pm\sqrt{-x^2+c} \,\,\,\left(\text{Simplify} \right)\)
\(\textbf{2)}\) \(\frac{dy}{dx}=y^2+4y\)
\(\,\,\,\,\, \frac{dy}{dx}=y^2+4y \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, \frac{dy}{dx}=y(y+4) \,\,\,\left(\text{Factoring}\right)\)
\(\,\,\,\,\, \frac{1}{y(y+4)}\frac{dy}{dx}=1\,\,\,\left(\text{Divide both sides by }y(y+4)\right)\)
\(\,\,\,\,\, \frac{1}{y(y+4)}dy=dx\,\,\,\left(\text{Multiply both sides by }dx\right)\)
\(\,\,\,\,\, \displaystyle \int\frac{1}{y(y+4)}dy= \int dx\,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, \displaystyle \int\frac{1}{y(y+4)}dy=x+c\,\,\,\left(\int 1 dx=x+c \right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle \frac{1}{y(y+4)}=\frac{A}{y}+\frac{B}{y+4}\,\,\,\left(\text{Setting up Partial Fraction Decomposition}\right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle \frac{y(y+4)}{y(y+4)}=\frac{Ay(y+4)}{y}+\frac{By(y+4)}{y+4}\,\,\,\left(\text{Multiply both sides by } y(y+4)\right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle 1=A(y+4)+By \,\,\,\left(\text{Simplify}\right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle 1=A(y+4)+By \,\,\,\left(\text{Simplify}\right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle 1=Ay+4A+By \,\,\,\left(\text{Distribute}\right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle 1=y(A+B)+4A \,\,\,\left(\text{Factor out }y\right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, 4A=1,/,A+B=0 \,\,\,\left(\text{Set corresponding parts equal}\right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, A=\frac{1}{4},\,A+B=0 \,\,\,\left(\text{Solve for }A\right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, A=\frac{1}{4},\,\frac{1}{4}+B=0 \,\,\,\left(\text{Plug in for }A\right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, A=\frac{1}{4},\,B=-\frac{1}{4} \,\,\,\left(\text{Solve for }B\right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle \frac{1}{y(y+4)}=\frac{1}{4y}-\frac{1}{4(y+4)}\,\,\,\left(\text{Finishing Partial Fraction Decomposition}\right)\)
\(\,\,\,\,\, \displaystyle \int\frac{1}{4y}-\frac{1}{4(y+4)}dy=x+c\,\,\,\left(\text{Substitute into the integral}\right)\)
\(\,\,\,\,\, \displaystyle \frac{1}{4} \int\frac{1}{y}-\frac{1}{(y+4)}dy=x+c\,\,\,\left(\text{Pull out the constant}\right)\)
\(\,\,\,\,\, \frac{1}{4}\left( \ln{|y|}-\ln{|y+4|}\right)=x+c\,\,\,\left(\int \frac{1}{y}=\ln{|y|}\right)\)
\(\,\,\,\,\, \frac{1}{4}\left( \ln{\left|\frac{y}{y+4}\right|}\right)=x+c\,\,\,\left( \ln{(a)}-\ln{(b)}=\ln{\left(\frac{a}{b}\right)} \right)\)
\(\,\,\,\,\, \left( \ln{\left|\frac{y}{y+4}\right|}\right)=4x+4c\,\,\,\left( \text{Multiply both sides by 4} \right)\)
\(\,\,\,\,\, \left( \ln{\left|\frac{y}{y+4}\right|}\right)=4x+c_1\,\,\,\left( \text{Let}4c=c_1 \right)\)
The answer is \( \ln{\left|\frac{y}{y+4}\right|}=4x+c_1 \)
\(\,\,\,\,\, \frac{dy}{dx}=y^2+4y \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, \frac{dy}{dx}=y(y+4) \,\,\,\left(\text{Factoring}\right)\)
\(\,\,\,\,\, \frac{1}{y(y+4)}\frac{dy}{dx}=1\,\,\,\left(\text{Divide both sides by }y(y+4)\right)\)
\(\,\,\,\,\, \frac{1}{y(y+4)}dy=dx\,\,\,\left(\text{Multiply both sides by }dx\right)\)
\(\,\,\,\,\, \displaystyle \int\frac{1}{y(y+4)}dy= \int dx\,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, \displaystyle \int\frac{1}{y(y+4)}dy=x+c\,\,\,\left(\int 1 dx=x+c \right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle \frac{1}{y(y+4)}=\frac{A}{y}+\frac{B}{y+4}\,\,\,\left(\text{Setting up Partial Fraction Decomposition}\right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle \frac{y(y+4)}{y(y+4)}=\frac{Ay(y+4)}{y}+\frac{By(y+4)}{y+4}\,\,\,\left(\text{Multiply both sides by } y(y+4)\right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle 1=A(y+4)+By \,\,\,\left(\text{Simplify}\right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle 1=A(y+4)+By \,\,\,\left(\text{Simplify}\right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle 1=Ay+4A+By \,\,\,\left(\text{Distribute}\right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle 1=y(A+B)+4A \,\,\,\left(\text{Factor out }y\right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, 4A=1,/,A+B=0 \,\,\,\left(\text{Set corresponding parts equal}\right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, A=\frac{1}{4},\,A+B=0 \,\,\,\left(\text{Solve for }A\right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, A=\frac{1}{4},\,\frac{1}{4}+B=0 \,\,\,\left(\text{Plug in for }A\right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, A=\frac{1}{4},\,B=-\frac{1}{4} \,\,\,\left(\text{Solve for }B\right)\)
\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle \frac{1}{y(y+4)}=\frac{1}{4y}-\frac{1}{4(y+4)}\,\,\,\left(\text{Finishing Partial Fraction Decomposition}\right)\)
\(\,\,\,\,\, \displaystyle \int\frac{1}{4y}-\frac{1}{4(y+4)}dy=x+c\,\,\,\left(\text{Substitute into the integral}\right)\)
\(\,\,\,\,\, \displaystyle \frac{1}{4} \int\frac{1}{y}-\frac{1}{(y+4)}dy=x+c\,\,\,\left(\text{Pull out the constant}\right)\)
\(\,\,\,\,\, \frac{1}{4}\left( \ln{|y|}-\ln{|y+4|}\right)=x+c\,\,\,\left(\int \frac{1}{y}=\ln{|y|}\right)\)
\(\,\,\,\,\, \frac{1}{4}\left( \ln{\left|\frac{y}{y+4}\right|}\right)=x+c\,\,\,\left( \ln{(a)}-\ln{(b)}=\ln{\left(\frac{a}{b}\right)} \right)\)
\(\,\,\,\,\, \left( \ln{\left|\frac{y}{y+4}\right|}\right)=4x+4c\,\,\,\left( \text{Multiply both sides by 4} \right)\)
\(\,\,\,\,\, \left( \ln{\left|\frac{y}{y+4}\right|}\right)=4x+c_1\,\,\,\left( \text{Let}4c=c_1 \right)\)
\(\textbf{3)}\) \(\frac{dy}{dx}=xe^y\)
\(\,\,\,\,\, \frac{dy}{dx}=xe^y \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, e^{-y}dy=x dx \,\,\,\left(\text{Multiply both sides by }dxe^{-y}\right)\)
\(\,\,\,\,\, \int e^{-y}dy= \int x dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, -e^{-y}= \frac{x^2}{2} +c \,\,\,\left(\text{Take the integrals} \right)\)
\(\,\,\,\,\, e^{-y}= -\frac{x^2}{2} -c \,\,\,\left(\text{Multiply both side by } \right)\)
\(\,\,\,\,\, \ln{e^{-y}}= \ln{\left(-\frac{x^2}{2} -c\right)} \,\,\,\left(\text{Natural log both sides } \right)\)
\(\,\,\,\,\, -y= \ln{\left(-\frac{x^2}{2} -c\right)} \,\,\,\left(\ln{e^x}=x \right)\)
\(\,\,\,\,\, -y= \ln{\left(-\frac{x^2}{2} +c_1\right)} \,\,\,\left(\text{Let }c_1=-c \right)\)
\(\,\,\,\,\, y= -\ln{\left(-\frac{x^2}{2} +c_1\right)} \,\,\,\left(\text{Multiply both sides by }-1 \right)\)
The answer is \( y= -\ln{\left(-\frac{x^2}{2} +c_1\right)} \)
\(\,\,\,\,\, \frac{dy}{dx}=xe^y \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, e^{-y}dy=x dx \,\,\,\left(\text{Multiply both sides by }dxe^{-y}\right)\)
\(\,\,\,\,\, \int e^{-y}dy= \int x dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, -e^{-y}= \frac{x^2}{2} +c \,\,\,\left(\text{Take the integrals} \right)\)
\(\,\,\,\,\, e^{-y}= -\frac{x^2}{2} -c \,\,\,\left(\text{Multiply both side by } \right)\)
\(\,\,\,\,\, \ln{e^{-y}}= \ln{\left(-\frac{x^2}{2} -c\right)} \,\,\,\left(\text{Natural log both sides } \right)\)
\(\,\,\,\,\, -y= \ln{\left(-\frac{x^2}{2} -c\right)} \,\,\,\left(\ln{e^x}=x \right)\)
\(\,\,\,\,\, -y= \ln{\left(-\frac{x^2}{2} +c_1\right)} \,\,\,\left(\text{Let }c_1=-c \right)\)
\(\,\,\,\,\, y= -\ln{\left(-\frac{x^2}{2} +c_1\right)} \,\,\,\left(\text{Multiply both sides by }-1 \right)\)
\(\textbf{4)}\) \(\frac{dy}{dx}=\frac{xy}{x^2+8}\)
\(\,\,\,\,\, \frac{dy}{dx}=\frac{xy}{x^2+8} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, \frac{1}{y}dy=\frac{x}{x^2+8}dx \,\,\,\left(\text{Multiply both sides by }\frac{1}{y}dx \right)\)
\(\,\,\,\,\, \int \frac{1}{y}dy=\int \frac{x}{x^2+8}dx \,\,\, \left(\text{Take the integral of both sides} \right)\)
\(\,\,\,\,\, \ln{|y|}=\int \frac{x}{x^2+8}dx \,\,\, \left(\text{Integrate left side} \right)\)
\(\,\,\,\,\, \ln{|y|}= \int \frac{x}{x^2+8}dx \,\,\, \left(\text{Pull out the constant }8 \right)\)
\(\,\,\,\,\,\,\,\,\,\, \text{Let }u=x^2+8 \)
\(\,\,\,\,\,\,\,\,\,\, \text{Then }du=2x dx \)
\(\,\,\,\,\,\,\,\,\,\, \text{Then }\frac{1}{2}du=x dx \)
\(\,\,\,\,\, \ln{|y|}=\frac{1}{2} \int \frac{1}{u}du \,\,\, \left(\text{Plug in }u \text{ and } du \right)\)
\(\,\,\,\,\, \ln{|y|}=\frac{1}{2} ln{|u|}+c \,\,\, \left(\text{Integrate }\right)\)
\(\,\,\,\,\, \ln{|y|}=\frac{1}{2} ln{|x^2+8|}+c \,\,\, \left(\text{plug in }x^2+8 \text{ for }u\right)\)
\(\,\,\,\,\, \ln{|y|}=\frac{1}{2} ln{|x^2+8|}+\frac{1}{2}\ln{c_1} \,\,\, \left(\text{Let }c=\frac{1}{2}ln{c_1} \right)\)
\(\,\,\,\,\, \ln{|y|}=\frac{1}{2} \left(ln{x^2+8}+\ln{c_1}\right) \,\,\, \left(\text{Factor out }\frac{1}{2} \right)\)
\(\,\,\,\,\, \ln{|y|}=\frac{1}{2} ln{c_1x^2+8} \,\,\, \left( \ln{a}+\ln{b}=\ln{ab} \right)\)
\(\,\,\,\,\, \ln{|y|}= ln{\left( c_1x^2+8 \right)^{1/2}} \,\,\, \left( c\ln{a}=\ln{a^c} \right)\)
\(\,\,\,\,\, \ln{|y|}= ln{\left( \sqrt{c_1\left(x^2+8\right)} \right)} \,\,\, \left( x^{1/2}=\sqrt{x} \right)\)
\(\,\,\,\,\, |y|= \sqrt{c_1\left(x^2+8\right)} \,\,\, \left( \ln{|x|}=\ln{y} \Rightarrow |x|=y \right)\)
\(\,\,\,\,\, y= \pm \sqrt{c_1\left(x^2+8\right) } \,\,\, \left( \ln{|x|}=\ln{y} \Rightarrow |x|=y \right)\)
\(\,\,\,\,\, y= \pm \sqrt{c_1} \cdot \sqrt{x^2+8} \,\,\, \left( \ln{|x|}=\ln{y} \Rightarrow |x|=y \right)\)
\(\,\,\,\,\, y= c_1 \sqrt{x^2+8} \,\,\, \left( \text{Let}c_1=\pm \sqrt{c_1} \right)\)
The answer is \( y= c_1 \sqrt{x^2+8} \)
\(\,\,\,\,\, \frac{dy}{dx}=\frac{xy}{x^2+8} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, \frac{1}{y}dy=\frac{x}{x^2+8}dx \,\,\,\left(\text{Multiply both sides by }\frac{1}{y}dx \right)\)
\(\,\,\,\,\, \int \frac{1}{y}dy=\int \frac{x}{x^2+8}dx \,\,\, \left(\text{Take the integral of both sides} \right)\)
\(\,\,\,\,\, \ln{|y|}=\int \frac{x}{x^2+8}dx \,\,\, \left(\text{Integrate left side} \right)\)
\(\,\,\,\,\, \ln{|y|}= \int \frac{x}{x^2+8}dx \,\,\, \left(\text{Pull out the constant }8 \right)\)
\(\,\,\,\,\,\,\,\,\,\, \text{Let }u=x^2+8 \)
\(\,\,\,\,\,\,\,\,\,\, \text{Then }du=2x dx \)
\(\,\,\,\,\,\,\,\,\,\, \text{Then }\frac{1}{2}du=x dx \)
\(\,\,\,\,\, \ln{|y|}=\frac{1}{2} \int \frac{1}{u}du \,\,\, \left(\text{Plug in }u \text{ and } du \right)\)
\(\,\,\,\,\, \ln{|y|}=\frac{1}{2} ln{|u|}+c \,\,\, \left(\text{Integrate }\right)\)
\(\,\,\,\,\, \ln{|y|}=\frac{1}{2} ln{|x^2+8|}+c \,\,\, \left(\text{plug in }x^2+8 \text{ for }u\right)\)
\(\,\,\,\,\, \ln{|y|}=\frac{1}{2} ln{|x^2+8|}+\frac{1}{2}\ln{c_1} \,\,\, \left(\text{Let }c=\frac{1}{2}ln{c_1} \right)\)
\(\,\,\,\,\, \ln{|y|}=\frac{1}{2} \left(ln{x^2+8}+\ln{c_1}\right) \,\,\, \left(\text{Factor out }\frac{1}{2} \right)\)
\(\,\,\,\,\, \ln{|y|}=\frac{1}{2} ln{c_1x^2+8} \,\,\, \left( \ln{a}+\ln{b}=\ln{ab} \right)\)
\(\,\,\,\,\, \ln{|y|}= ln{\left( c_1x^2+8 \right)^{1/2}} \,\,\, \left( c\ln{a}=\ln{a^c} \right)\)
\(\,\,\,\,\, \ln{|y|}= ln{\left( \sqrt{c_1\left(x^2+8\right)} \right)} \,\,\, \left( x^{1/2}=\sqrt{x} \right)\)
\(\,\,\,\,\, |y|= \sqrt{c_1\left(x^2+8\right)} \,\,\, \left( \ln{|x|}=\ln{y} \Rightarrow |x|=y \right)\)
\(\,\,\,\,\, y= \pm \sqrt{c_1\left(x^2+8\right) } \,\,\, \left( \ln{|x|}=\ln{y} \Rightarrow |x|=y \right)\)
\(\,\,\,\,\, y= \pm \sqrt{c_1} \cdot \sqrt{x^2+8} \,\,\, \left( \ln{|x|}=\ln{y} \Rightarrow |x|=y \right)\)
\(\,\,\,\,\, y= c_1 \sqrt{x^2+8} \,\,\, \left( \text{Let}c_1=\pm \sqrt{c_1} \right)\)
\(\textbf{5)}\) \(\frac{dy}{dx}=\frac{e^x}{1+e^x}\)
\(\,\,\,\,\, \frac{dy}{dx}=\frac{e^x}{1+e^x} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, dy=\frac{e^x}{1+e^x} dx \,\,\,\left(\text{Multiply both sides by } dx\right)\)
\(\,\,\,\,\, \int dy= \int \frac{e^x}{1+e^x} dx \,\,\,\left(\text{Take integral of both sides }\right)\)
\(\,\,\,\,\, y= \int \frac{e^x}{1+e^x} dx \,\,\,\left( \int dy=y\right)\)
\(\,\,\,\,\,\,\,\,\,\, u=1+e^x,\)
\(\,\,\,\,\,\,\,\,\,\, du=e^x dx,\)
\(\,\,\,\,\, y= \int \frac{1}{u} du \,\,\,\left(\text{Plug in }u \text{ and } du \right)\)
\(\,\,\,\,\, y= \ln{|u|}+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y= \ln{\left(1+e^x\right)}+c \,\,\,\left(\text{Plug in }1+e^x \text{ for } u\right)\)
The answer is \( y= \ln{\left(1+e^x\right)}+c \)
\(\,\,\,\,\, \frac{dy}{dx}=\frac{e^x}{1+e^x} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, dy=\frac{e^x}{1+e^x} dx \,\,\,\left(\text{Multiply both sides by } dx\right)\)
\(\,\,\,\,\, \int dy= \int \frac{e^x}{1+e^x} dx \,\,\,\left(\text{Take integral of both sides }\right)\)
\(\,\,\,\,\, y= \int \frac{e^x}{1+e^x} dx \,\,\,\left( \int dy=y\right)\)
\(\,\,\,\,\,\,\,\,\,\, u=1+e^x,\)
\(\,\,\,\,\,\,\,\,\,\, du=e^x dx,\)
\(\,\,\,\,\, y= \int \frac{1}{u} du \,\,\,\left(\text{Plug in }u \text{ and } du \right)\)
\(\,\,\,\,\, y= \ln{|u|}+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y= \ln{\left(1+e^x\right)}+c \,\,\,\left(\text{Plug in }1+e^x \text{ for } u\right)\)
\(\textbf{6)}\) \(\frac{dy}{dx}=10xy^2\)
\(\,\,\,\,\, \frac{dy}{dx}=10xy^2 \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, y^{-2} dy=10x dx \,\,\,\left(\text{Multiply both sides by } y^{-2} dx\right)\)
\(\,\,\,\,\, \int y^{-2} dy=\int 10x dx \,\,\,\left(\text{Take the integral of both sides}\right)\)
\(\,\,\,\,\, -y^{-1} = \frac{10x^2}{2}+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y^{-1} = -5x^2-c \,\,\,\left(\text{Simplify and divide by} -1\right)\)
\(\,\,\,\,\, \left(y^{-1}\right)^{-1} = \left(-5x^2+c\right)^{-1} \,\,\,\left(\text{Take both sides to the} -1 \text{ power }\right)\)
\(\,\,\,\,\, y = \frac{1}{-5x^2-c} \,\,\,\left(\text{Simplify}\right)\)
\(\,\,\,\,\, y = \frac{1}{-5x^2+c} \,\,\,\left(\text{Let}c=-c\right)\)
The answer is \( y = \frac{1}{-5x^2+c} \)
\(\,\,\,\,\, \frac{dy}{dx}=10xy^2 \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, y^{-2} dy=10x dx \,\,\,\left(\text{Multiply both sides by } y^{-2} dx\right)\)
\(\,\,\,\,\, \int y^{-2} dy=\int 10x dx \,\,\,\left(\text{Take the integral of both sides}\right)\)
\(\,\,\,\,\, -y^{-1} = \frac{10x^2}{2}+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y^{-1} = -5x^2-c \,\,\,\left(\text{Simplify and divide by} -1\right)\)
\(\,\,\,\,\, \left(y^{-1}\right)^{-1} = \left(-5x^2+c\right)^{-1} \,\,\,\left(\text{Take both sides to the} -1 \text{ power }\right)\)
\(\,\,\,\,\, y = \frac{1}{-5x^2-c} \,\,\,\left(\text{Simplify}\right)\)
\(\,\,\,\,\, y = \frac{1}{-5x^2+c} \,\,\,\left(\text{Let}c=-c\right)\)
\(\textbf{7)}\) \(\frac{dy}{dx} = \frac{2x}{y}\)
\(\,\,\,\,\, \frac{dy}{dx} = \frac{2x}{y} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, y \, dy = 2x \, dx \,\,\,\left(\text{Multiply both sides by } y \, dx\right)\)
\(\,\,\,\,\, \int y \, dy = \int 2x \, dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, \frac{y^2}{2} = x^2 + c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y^2 = 2x^2 + 2c \,\,\,\left(\text{Multiply both sides by } 2\right)\)
\(\,\,\,\,\, y^2 = 2x^2 + C \,\,\,\left(\text{Let } C = 2c\right)\)
\(\,\,\,\,\, y = \pm\sqrt{2x^2 + C} \,\,\,\left(\text{Take the square root of both sides}\right)\)
The answer is \( y = \pm\sqrt{2x^2 + C} \)
\(\,\,\,\,\, \frac{dy}{dx} = \frac{2x}{y} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, y \, dy = 2x \, dx \,\,\,\left(\text{Multiply both sides by } y \, dx\right)\)
\(\,\,\,\,\, \int y \, dy = \int 2x \, dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, \frac{y^2}{2} = x^2 + c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y^2 = 2x^2 + 2c \,\,\,\left(\text{Multiply both sides by } 2\right)\)
\(\,\,\,\,\, y^2 = 2x^2 + C \,\,\,\left(\text{Let } C = 2c\right)\)
\(\,\,\,\,\, y = \pm\sqrt{2x^2 + C} \,\,\,\left(\text{Take the square root of both sides}\right)\)
\(\textbf{8)}\) \(\frac{dy}{dx} = y \ln(y)\)
\(\,\,\,\,\, \frac{dy}{dx} = y \ln(y) \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, \frac{1}{y \ln(y)} \, dy = dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int \frac{1}{y \ln(y)} \, dy = \int dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, \int \frac{1}{y \ln(y)} \, dy = x + c \,\,\,\left(\text{Integrate right side}\right)\)
\(\,\,\,\,\, \text{Let } u = \ln(y), \,\,\, du = \frac{1}{y} \, dy \,\,\,\left(\text{Substitute}\right)\)
\(\,\,\,\,\, \int \frac{1}{u} \, du = x + c \,\,\,\left(\text{Substitute and integrate}\right)\)
\(\,\,\,\,\, \ln|u| = x + c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, \ln|\ln(y)| = x + c \,\,\,\left(\text{Substitute back for } u\right)\)
\(\,\,\,\,\, \ln(y) = \pm e^{x+c} \,\,\,\left(\text{Exponentiate both sides}\right)\)
\(\,\,\,\,\, y = e^{\pm e^{x+c}} \,\,\,\left(\text{Exponentiate again}\right)\)
The answer is \( y = e^{\pm e^{x+c}} \)
\(\,\,\,\,\, \frac{dy}{dx} = y \ln(y) \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, \frac{1}{y \ln(y)} \, dy = dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int \frac{1}{y \ln(y)} \, dy = \int dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, \int \frac{1}{y \ln(y)} \, dy = x + c \,\,\,\left(\text{Integrate right side}\right)\)
\(\,\,\,\,\, \text{Let } u = \ln(y), \,\,\, du = \frac{1}{y} \, dy \,\,\,\left(\text{Substitute}\right)\)
\(\,\,\,\,\, \int \frac{1}{u} \, du = x + c \,\,\,\left(\text{Substitute and integrate}\right)\)
\(\,\,\,\,\, \ln|u| = x + c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, \ln|\ln(y)| = x + c \,\,\,\left(\text{Substitute back for } u\right)\)
\(\,\,\,\,\, \ln(y) = \pm e^{x+c} \,\,\,\left(\text{Exponentiate both sides}\right)\)
\(\,\,\,\,\, y = e^{\pm e^{x+c}} \,\,\,\left(\text{Exponentiate again}\right)\)
\(\textbf{9)}\) \(\frac{dy}{dx}=3xy\)
\(\,\,\,\,\, \frac{dy}{dx}=3xy \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, \frac{1}{y}dy=3x dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int \frac{1}{y}dy=\int 3x dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, \ln{|y|}=\frac{3x^2}{2}+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y=ce^{\frac{3x^2}{2}} \,\,\,\left(\text{Solve for }y\right)\)
The answer is \( y=ce^{\frac{3x^2}{2}} \)
\(\,\,\,\,\, \frac{dy}{dx}=3xy \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, \frac{1}{y}dy=3x dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int \frac{1}{y}dy=\int 3x dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, \ln{|y|}=\frac{3x^2}{2}+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y=ce^{\frac{3x^2}{2}} \,\,\,\left(\text{Solve for }y\right)\)
\(\textbf{10)}\) \(\frac{dy}{dx}=\frac{x}{y+2}\)
\(\,\,\,\,\, \frac{dy}{dx}=\frac{x}{y+2} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, (y+2)dy=x dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int (y+2)dy=\int x dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, \frac{y^2}{2}+2y=\frac{x^2}{2}+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y^2+4y=x^2+c \,\,\,\left(\text{Multiply both sides by }2\right)\)
\(\,\,\,\,\, (y+2)^2=x^2+c \,\,\,\left(\text{Complete the square and absorb constants}\right)\)
\(\,\,\,\,\, y=-2\pm\sqrt{x^2+c} \,\,\,\left(\text{Solve for }y\right)\)
The answer is \( y=-2\pm\sqrt{x^2+c} \)
\(\,\,\,\,\, \frac{dy}{dx}=\frac{x}{y+2} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, (y+2)dy=x dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int (y+2)dy=\int x dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, \frac{y^2}{2}+2y=\frac{x^2}{2}+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y^2+4y=x^2+c \,\,\,\left(\text{Multiply both sides by }2\right)\)
\(\,\,\,\,\, (y+2)^2=x^2+c \,\,\,\left(\text{Complete the square and absorb constants}\right)\)
\(\,\,\,\,\, y=-2\pm\sqrt{x^2+c} \,\,\,\left(\text{Solve for }y\right)\)
\(\textbf{11)}\) \(\frac{dy}{dx}=\frac{y}{x}\)
\(\,\,\,\,\, \frac{dy}{dx}=\frac{y}{x} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, \frac{1}{y}dy=\frac{1}{x}dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int \frac{1}{y}dy=\int \frac{1}{x}dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, \ln{|y|}=\ln{|x|}+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y=cx \,\,\,\left(\text{Solve for }y\right)\)
The answer is \( y=cx \)
\(\,\,\,\,\, \frac{dy}{dx}=\frac{y}{x} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, \frac{1}{y}dy=\frac{1}{x}dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int \frac{1}{y}dy=\int \frac{1}{x}dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, \ln{|y|}=\ln{|x|}+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y=cx \,\,\,\left(\text{Solve for }y\right)\)
\(\textbf{12)}\) \(\frac{dy}{dx}=6x^2e^{-y}\)
\(\,\,\,\,\, \frac{dy}{dx}=6x^2e^{-y} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, e^y dy=6x^2 dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int e^y dy=\int 6x^2 dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, e^y=2x^3+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y=\ln{\left(2x^3+c\right)} \,\,\,\left(\text{Natural log both sides}\right)\)
The answer is \( y=\ln{\left(2x^3+c\right)} \)
\(\,\,\,\,\, \frac{dy}{dx}=6x^2e^{-y} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, e^y dy=6x^2 dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int e^y dy=\int 6x^2 dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, e^y=2x^3+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y=\ln{\left(2x^3+c\right)} \,\,\,\left(\text{Natural log both sides}\right)\)
\(\textbf{13)}\) \(\frac{dy}{dx}=(x+1)y^2\)
\(\,\,\,\,\, \frac{dy}{dx}=(x+1)y^2 \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, y^{-2}dy=(x+1)dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int y^{-2}dy=\int (x+1)dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, -y^{-1}=\frac{x^2}{2}+x+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y=\frac{1}{-\frac{x^2}{2}-x+c} \,\,\,\left(\text{Solve for }y\right)\)
The answer is \( y=\frac{1}{-\frac{x^2}{2}-x+c} \)
\(\,\,\,\,\, \frac{dy}{dx}=(x+1)y^2 \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, y^{-2}dy=(x+1)dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int y^{-2}dy=\int (x+1)dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, -y^{-1}=\frac{x^2}{2}+x+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y=\frac{1}{-\frac{x^2}{2}-x+c} \,\,\,\left(\text{Solve for }y\right)\)
\(\textbf{14)}\) \(\frac{dy}{dx}=\frac{\cos{x}}{y}\)
\(\,\,\,\,\, \frac{dy}{dx}=\frac{\cos{x}}{y} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, y dy=\cos{x} dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int y dy=\int \cos{x} dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, \frac{y^2}{2}=\sin{x}+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y=\pm\sqrt{2\sin{x}+c} \,\,\,\left(\text{Solve for }y\right)\)
The answer is \( y=\pm\sqrt{2\sin{x}+c} \)
\(\,\,\,\,\, \frac{dy}{dx}=\frac{\cos{x}}{y} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, y dy=\cos{x} dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int y dy=\int \cos{x} dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, \frac{y^2}{2}=\sin{x}+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y=\pm\sqrt{2\sin{x}+c} \,\,\,\left(\text{Solve for }y\right)\)
\(\textbf{15)}\) \(\frac{dy}{dx}=\frac{4x}{1+y^2}\)
\(\,\,\,\,\, \frac{dy}{dx}=\frac{4x}{1+y^2} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, (1+y^2)dy=4x dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int (1+y^2)dy=\int 4x dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, y+\frac{y^3}{3}=2x^2+c \,\,\,\left(\text{Integrate}\right)\)
The answer is \( y+\frac{y^3}{3}=2x^2+c \)
\(\,\,\,\,\, \frac{dy}{dx}=\frac{4x}{1+y^2} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, (1+y^2)dy=4x dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int (1+y^2)dy=\int 4x dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, y+\frac{y^3}{3}=2x^2+c \,\,\,\left(\text{Integrate}\right)\)
\(\textbf{16)}\) \(\frac{dy}{dx}=\frac{2x+3}{2y}\)
\(\,\,\,\,\, \frac{dy}{dx}=\frac{2x+3}{2y} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, 2y dy=(2x+3)dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int 2y dy=\int (2x+3)dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, y^2=x^2+3x+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y=\pm\sqrt{x^2+3x+c} \,\,\,\left(\text{Solve for }y\right)\)
The answer is \( y=\pm\sqrt{x^2+3x+c} \)
\(\,\,\,\,\, \frac{dy}{dx}=\frac{2x+3}{2y} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, 2y dy=(2x+3)dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int 2y dy=\int (2x+3)dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, y^2=x^2+3x+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y=\pm\sqrt{x^2+3x+c} \,\,\,\left(\text{Solve for }y\right)\)
\(\textbf{17)}\) \(\frac{dy}{dx}=e^{x-y}\)
\(\,\,\,\,\, \frac{dy}{dx}=e^{x-y} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, \frac{dy}{dx}=e^xe^{-y} \,\,\,\left(\text{Rewrite}\right)\)
\(\,\,\,\,\, e^y dy=e^x dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int e^y dy=\int e^x dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, e^y=e^x+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y=\ln{\left(e^x+c\right)} \,\,\,\left(\text{Natural log both sides}\right)\)
The answer is \( y=\ln{\left(e^x+c\right)} \)
\(\,\,\,\,\, \frac{dy}{dx}=e^{x-y} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, \frac{dy}{dx}=e^xe^{-y} \,\,\,\left(\text{Rewrite}\right)\)
\(\,\,\,\,\, e^y dy=e^x dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int e^y dy=\int e^x dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, e^y=e^x+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y=\ln{\left(e^x+c\right)} \,\,\,\left(\text{Natural log both sides}\right)\)
\(\textbf{18)}\) \(\frac{dy}{dx}=\frac{y}{x^2+1}\)
\(\,\,\,\,\, \frac{dy}{dx}=\frac{y}{x^2+1} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, \frac{1}{y}dy=\frac{1}{x^2+1}dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int \frac{1}{y}dy=\int \frac{1}{x^2+1}dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, \ln{|y|}=\tan^{-1}{x}+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y=ce^{\tan^{-1}{x}} \,\,\,\left(\text{Solve for }y\right)\)
The answer is \( y=ce^{\tan^{-1}{x}} \)
\(\,\,\,\,\, \frac{dy}{dx}=\frac{y}{x^2+1} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, \frac{1}{y}dy=\frac{1}{x^2+1}dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int \frac{1}{y}dy=\int \frac{1}{x^2+1}dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, \ln{|y|}=\tan^{-1}{x}+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y=ce^{\tan^{-1}{x}} \,\,\,\left(\text{Solve for }y\right)\)
\(\textbf{19)}\) \(\frac{dy}{dx}=x\sqrt{y}\)
\(\,\,\,\,\, \frac{dy}{dx}=x\sqrt{y} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, y^{-\frac{1}{2}}dy=x dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int y^{-\frac{1}{2}}dy=\int x dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, 2\sqrt{y}=\frac{x^2}{2}+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, \sqrt{y}=\frac{x^2}{4}+c \,\,\,\left(\text{Divide by }2\text{ and absorb constants}\right)\)
\(\,\,\,\,\, y=\left(\frac{x^2}{4}+c\right)^2 \,\,\,\left(\text{Square both sides}\right)\)
The answer is \( y=\left(\frac{x^2}{4}+c\right)^2 \)
\(\,\,\,\,\, \frac{dy}{dx}=x\sqrt{y} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, y^{-\frac{1}{2}}dy=x dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int y^{-\frac{1}{2}}dy=\int x dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, 2\sqrt{y}=\frac{x^2}{2}+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, \sqrt{y}=\frac{x^2}{4}+c \,\,\,\left(\text{Divide by }2\text{ and absorb constants}\right)\)
\(\,\,\,\,\, y=\left(\frac{x^2}{4}+c\right)^2 \,\,\,\left(\text{Square both sides}\right)\)
\(\textbf{20)}\) \(\frac{dy}{dx}=\frac{y+1}{x+2}\)
\(\,\,\,\,\, \frac{dy}{dx}=\frac{y+1}{x+2} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, \frac{1}{y+1}dy=\frac{1}{x+2}dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int \frac{1}{y+1}dy=\int \frac{1}{x+2}dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, \ln{|y+1|}=\ln{|x+2|}+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y+1=c(x+2) \,\,\,\left(\text{Exponentiate both sides}\right)\)
\(\,\,\,\,\, y=c(x+2)-1 \,\,\,\left(\text{Solve for }y\right)\)
The answer is \( y=c(x+2)-1 \)
\(\,\,\,\,\, \frac{dy}{dx}=\frac{y+1}{x+2} \,\,\,\left(\text{Given}\right)\)
\(\,\,\,\,\, \frac{1}{y+1}dy=\frac{1}{x+2}dx \,\,\,\left(\text{Separate variables}\right)\)
\(\,\,\,\,\, \int \frac{1}{y+1}dy=\int \frac{1}{x+2}dx \,\,\,\left(\text{Integrate both sides}\right)\)
\(\,\,\,\,\, \ln{|y+1|}=\ln{|x+2|}+c \,\,\,\left(\text{Integrate}\right)\)
\(\,\,\,\,\, y+1=c(x+2) \,\,\,\left(\text{Exponentiate both sides}\right)\)
\(\,\,\,\,\, y=c(x+2)-1 \,\,\,\left(\text{Solve for }y\right)\)
See Related Pages\(\)
\(\bullet \text{ Differential Equations Calculator (Symbolab)}\)
\(\bullet\text{ Indefinite Integrals (Power Rule)}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int x^4-2x^2+5x \,dx…\)
\(\bullet\text{ Andymath Homepage}\)
