1st Order Differential Equations (Separable)

Practice Problems

\(\textbf{1)}\) \(\frac{dy}{dx}=-\frac{x}{y}\)

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\(\,\,\,\,\, \frac{dy}{dx}=-\frac{x}{y} \,\,\,\left(\text{Given}\right)\)

\(\,\,\,\,\, y dy=-x dx \,\,\,\left(\text{Multiply both sides by } y\, dx\right)\)

\(\,\,\,\,\, \int y dy= \int -x dx \,\,\,\left(\text{Take integral of both sides}\right)\)

\(\,\,\,\,\, \frac{y^2}{2} = \frac{-x^2}{2}+c \,\,\,\left(\text{Integrate}\right)\)

\(\,\,\,\,\, y^2 = -x^2+2c \,\,\,\left(\text{Multiply both sides by }2 \right)\)

\(\,\,\,\,\, y^2 = -x^2+c \,\,\,\left(\text{Let }c=2c \right)\)

\(\,\,\,\,\, \sqrt{y^2} = \sqrt{-x^2+c} \,\,\,\left(\text{Square root both sides } \right)\)

\(\,\,\,\,\, y = \pm\sqrt{-x^2+c} \,\,\,\left(\text{Simplify} \right)\)

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The answer is \( y = \pm\sqrt{-x^2+c} \)

\(\textbf{2)}\) \(\frac{dy}{dx}=y^2+4y\)

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\(\,\,\,\,\, \frac{dy}{dx}=y^2+4y \,\,\,\left(\text{Given}\right)\)

\(\,\,\,\,\, \frac{dy}{dx}=y(y+4) \,\,\,\left(\text{Factoring}\right)\)

\(\,\,\,\,\, \frac{1}{y(y+4)}\frac{dy}{dx}=1\,\,\,\left(\text{Divide both sides by }y(y+4)\right)\)

\(\,\,\,\,\, \frac{1}{y(y+4)}dy=dx\,\,\,\left(\text{Multiply both sides by }dx\right)\)

\(\,\,\,\,\, \displaystyle \int\frac{1}{y(y+4)}dy= \int dx\,\,\,\left(\text{Integrate both sides}\right)\)

\(\,\,\,\,\, \displaystyle \int\frac{1}{y(y+4)}dy=x+c\,\,\,\left(\int 1 dx=x+c \right)\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle \frac{1}{y(y+4)}=\frac{A}{y}+\frac{B}{y+4}\,\,\,\left(\text{Setting up Partial Fraction Decomposition}\right)\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle \frac{y(y+4)}{y(y+4)}=\frac{Ay(y+4)}{y}+\frac{By(y+4)}{y+4}\,\,\,\left(\text{Multiply both sides by } y(y+4)\right)\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle 1=A(y+4)+By \,\,\,\left(\text{Simplify}\right)\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle 1=A(y+4)+By \,\,\,\left(\text{Simplify}\right)\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle 1=Ay+4A+By \,\,\,\left(\text{Distribute}\right)\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle 1=y(A+B)+4A \,\,\,\left(\text{Factor out }y\right)\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, 4A=1,/,A+B=0 \,\,\,\left(\text{Set corresponding parts equal}\right)\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, A=\frac{1}{4},\,A+B=0 \,\,\,\left(\text{Solve for }A\right)\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, A=\frac{1}{4},\,\frac{1}{4}+B=0 \,\,\,\left(\text{Plug in for }A\right)\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, A=\frac{1}{4},\,B=-\frac{1}{4} \,\,\,\left(\text{Solve for }B\right)\)

\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \displaystyle \frac{1}{y(y+4)}=\frac{1}{4y}-\frac{1}{4(y+4)}\,\,\,\left(\text{Finishing Partial Fraction Decomposition}\right)\)

\(\,\,\,\,\, \displaystyle \int\frac{1}{4y}-\frac{1}{4(y+4)}dy=x+c\,\,\,\left(\text{Substitute into the integral}\right)\)

\(\,\,\,\,\, \displaystyle \frac{1}{4} \int\frac{1}{y}-\frac{1}{(y+4)}dy=x+c\,\,\,\left(\text{Pull out the constant}\right)\)

\(\,\,\,\,\, \frac{1}{4}\left( \ln{|y|}-\ln{|y+4|}\right)=x+c\,\,\,\left(\int \frac{1}{y}=\ln{|y|}\right)\)

\(\,\,\,\,\, \frac{1}{4}\left( \ln{\left|\frac{y}{y+4}\right|}\right)=x+c\,\,\,\left( \ln{(a)}-\ln{(b)}=\ln{\left(\frac{a}{b}\right)} \right)\)

\(\,\,\,\,\, \left( \ln{\left|\frac{y}{y+4}\right|}\right)=4x+4c\,\,\,\left( \text{Multiply both sides by 4} \right)\)

\(\,\,\,\,\, \left( \ln{\left|\frac{y}{y+4}\right|}\right)=4x+c_1\,\,\,\left( \text{Let}4c=c_1 \right)\)

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The answer is \( \ln{\left|\frac{y}{y+4}\right|}=4x+c_1 \)

\(\textbf{3)}\) \(\frac{dy}{dx}=xe^y\)

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\(\,\,\,\,\, \frac{dy}{dx}=xe^y \,\,\,\left(\text{Given}\right)\)

\(\,\,\,\,\, e^{-y}dy=x dx \,\,\,\left(\text{Multiply both sides by }dxe^{-y}\right)\)

\(\,\,\,\,\, \int e^{-y}dy= \int x dx \,\,\,\left(\text{Integrate both sides}\right)\)

\(\,\,\,\,\, -e^{-y}= \frac{x^2}{2} +c \,\,\,\left(\text{Take the integrals} \right)\)

\(\,\,\,\,\, e^{-y}= -\frac{x^2}{2} -c \,\,\,\left(\text{Multiply both side by } \right)\)

\(\,\,\,\,\, \ln{e^{-y}}= \ln{\left(-\frac{x^2}{2} -c\right)} \,\,\,\left(\text{Natural log both sides } \right)\)

\(\,\,\,\,\, -y= \ln{\left(-\frac{x^2}{2} -c\right)} \,\,\,\left(\ln{e^x}=x \right)\)

\(\,\,\,\,\, -y= \ln{\left(-\frac{x^2}{2} +c_1\right)} \,\,\,\left(\text{Let }c_1=-c \right)\)

\(\,\,\,\,\, y= -\ln{\left(-\frac{x^2}{2} +c_1\right)} \,\,\,\left(\text{Multiply both sides by }-1 \right)\)

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The answer is \( y= -\ln{\left(-\frac{x^2}{2} +c_1\right)} \)

\(\textbf{4)}\) \(\frac{dy}{dx}=\frac{xy}{x^2+8}\)

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\(\,\,\,\,\, \frac{dy}{dx}=\frac{xy}{x^2+8} \,\,\,\left(\text{Given}\right)\)

\(\,\,\,\,\, \frac{1}{y}dy=\frac{x}{x^2+8}dx \,\,\,\left(\text{Multiply both sides by }\frac{1}{y}dx \right)\)

\(\,\,\,\,\, \int \frac{1}{y}dy=\int \frac{x}{x^2+8}dx \,\,\, \left(\text{Take the integral of both sides} \right)\)

\(\,\,\,\,\, \ln{|y|}=\int \frac{x}{x^2+8}dx \,\,\, \left(\text{Integrate left side} \right)\)

\(\,\,\,\,\, \ln{|y|}= \int \frac{x}{x^2+8}dx \,\,\, \left(\text{Pull out the constant }8 \right)\)

\(\,\,\,\,\,\,\,\,\,\, \text{Let }u=x^2+8 \)

\(\,\,\,\,\,\,\,\,\,\, \text{Then }du=2x dx \)

\(\,\,\,\,\,\,\,\,\,\, \text{Then }\frac{1}{2}du=x dx \)

\(\,\,\,\,\, \ln{|y|}=\frac{1}{2} \int \frac{1}{u}du \,\,\, \left(\text{Plug in }u \text{ and } du \right)\)

\(\,\,\,\,\, \ln{|y|}=\frac{1}{2} ln{|u|}+c \,\,\, \left(\text{Integrate }\right)\)

\(\,\,\,\,\, \ln{|y|}=\frac{1}{2} ln{|x^2+8|}+c \,\,\, \left(\text{plug in }x^2+8 \text{ for }u\right)\)

\(\,\,\,\,\, \ln{|y|}=\frac{1}{2} ln{|x^2+8|}+\frac{1}{2}\ln{c_1} \,\,\, \left(\text{Let }c=\frac{1}{2}ln{c_1} \right)\)

\(\,\,\,\,\, \ln{|y|}=\frac{1}{2} \left(ln{x^2+8}+\ln{c_1}\right) \,\,\, \left(\text{Factor out }\frac{1}{2} \right)\)

\(\,\,\,\,\, \ln{|y|}=\frac{1}{2} ln{c_1x^2+8} \,\,\, \left( \ln{a}+\ln{b}=\ln{ab} \right)\)

\(\,\,\,\,\, \ln{|y|}= ln{\left( c_1x^2+8 \right)^{1/2}} \,\,\, \left( c\ln{a}=\ln{a^c} \right)\)

\(\,\,\,\,\, \ln{|y|}= ln{\left( \sqrt{c_1\left(x^2+8\right)} \right)} \,\,\, \left( x^{1/2}=\sqrt{x} \right)\)

\(\,\,\,\,\, |y|= \sqrt{c_1\left(x^2+8\right)} \,\,\, \left( \ln{|x|}=\ln{y} \Rightarrow |x|=y \right)\)

\(\,\,\,\,\, y= \pm \sqrt{c_1\left(x^2+8\right) } \,\,\, \left( \ln{|x|}=\ln{y} \Rightarrow |x|=y \right)\)

\(\,\,\,\,\, y= \pm \sqrt{c_1} \cdot \sqrt{x^2+8} \,\,\, \left( \ln{|x|}=\ln{y} \Rightarrow |x|=y \right)\)

\(\,\,\,\,\, y= c_1 \sqrt{x^2+8} \,\,\, \left( \text{Let}c_1=\pm \sqrt{c_1} \right)\)

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The answer is \( y= c_1 \sqrt{x^2+8} \)

\(\textbf{5)}\) \(\frac{dy}{dx}=\frac{e^x}{1+e^x}\)

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\(\,\,\,\,\, \frac{dy}{dx}=\frac{e^x}{1+e^x} \,\,\,\left(\text{Given}\right)\)

\(\,\,\,\,\, dy=\frac{e^x}{1+e^x} dx \,\,\,\left(\text{Multiply both sides by } dx\right)\)

\(\,\,\,\,\, \int dy= \int \frac{e^x}{1+e^x} dx \,\,\,\left(\text{Take integral of both sides }\right)\)

\(\,\,\,\,\, y= \int \frac{e^x}{1+e^x} dx \,\,\,\left( \int dy=y\right)\)

\(\,\,\,\,\,\,\,\,\,\, u=1+e^x,\)

\(\,\,\,\,\,\,\,\,\,\, du=e^x dx,\)

\(\,\,\,\,\, y= \int \frac{1}{u} du \,\,\,\left(\text{Plug in }u \text{ and } du \right)\)

\(\,\,\,\,\, y= \ln{|u|}+c \,\,\,\left(\text{Integrate}\right)\)

\(\,\,\,\,\, y= \ln{\left(1+e^x\right)}+c \,\,\,\left(\text{Plug in }1+e^x \text{ for } u\right)\)

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The answer is \( y= \ln{\left(1+e^x\right)}+c \)

\(\textbf{6)}\) \(\frac{dy}{dx}=10xy^2\)

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\(\,\,\,\,\, \frac{dy}{dx}=10xy^2 \,\,\,\left(\text{Given}\right)\)

\(\,\,\,\,\, y^{-2} dy=10x dx \,\,\,\left(\text{Multiply both sides by } y^{-2} dx\right)\)

\(\,\,\,\,\, \int y^{-2} dy=\int 10x dx \,\,\,\left(\text{Take the integral of both sides}\right)\)

\(\,\,\,\,\, -y^{-1} = \frac{10x^2}{2}+c \,\,\,\left(\text{Integrate}\right)\)

\(\,\,\,\,\, y^{-1} = -5x^2-c \,\,\,\left(\text{Simplify and divide by} -1\right)\)

\(\,\,\,\,\, \left(y^{-1}\right)^{-1} = \left(-5x^2+c\right)^{-1} \,\,\,\left(\text{Take both sides to the} -1 \text{ power }\right)\)

\(\,\,\,\,\, y = \frac{1}{-5x^2-c} \,\,\,\left(\text{Simplify}\right)\)

\(\,\,\,\,\, y = \frac{1}{-5x^2+c} \,\,\,\left(\text{Let}c=-c\right)\)

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The answer is \( y = \frac{1}{-5x^2+c} \)

See Related Pages\(\)

\(\bullet\text{ Indefinite Integrals (Power Rule)}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int x^4-2x^2+5x \,dx…\)

\(\bullet\text{ Andymath Homepage}\)

In Summary

First order differential equations involve a single derivative of a function and can be used to model a wide range of phenomena in fields such as physics, engineering, and economics. One type of first order differential equation is a separable equation, which can be written in the form \(\frac{dy}{dx} = f(x)g(y)\). These types of equations can be solved by separating the variables and integrating both sides of the equation.

To solve a separable differential equation, the first step is to rearrange the equation so that all the terms involving the derivative of the function are on one side and all the other terms are on the other side. The next step is to divide both sides of the equation by the expression involving the derivative. This will result in an equation of the form \(g(y)dy = f(x)dx\).

At this point, the equation can be solved by separating the variables and integrating both sides. Once both sides of the equation have been integrated, the constants of integration can be determined by applying any conditions that are given. The final solution for the differential equation will be a function that satisfies the original equation and the boundary conditions.

Differential equations are an important tool for modeling and understanding many real-world phenomena. They are used in fields such as physics, engineering, and economics to describe the behavior of systems that change over time.

Differential equations involve the derivative of a function, which represents the rate of change of the function. By solving differential equations, we can find functions that describe how a system changes over time, which can provide insight into the underlying processes and behaviors of the system.

There are different types of differential equations, including first order, second order, and higher order equations, each of which involves a different number of derivatives. Different techniques are used to solve these different types of equations, depending on the form of the equation and the specific problem being addressed.

Overall, the study of differential equations is an important area of mathematics that has many practical applications and is essential for understanding and analyzing many complex systems in the real world.

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