This page contains a mixed bag of practice problems solving first order differential equations based on a video from one of our favorite instructors. We have laid out each practice problem and included the video clip containing each solution.
We recommend that you download this pdf of his list of problems before starting. However, his videos do not follow this pdf exactly.
Make sure you support the guy that did this video. He put a LOT of work into, not just doing the video, but also preparing the problems and making sure his solutions were correct. He did a GREAT job. So go to YouTube and like this video and follow him. He is one of our favorite instructors. (By the way, we are not receiving any compensation from him. We just think his videos will help you.) Here is his YouTube channel link.
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Practice
Find the general solution to these differential equations. If initial conditions are given, find the particular solution as well.
Basic
\( (\sin(y)  y\sin(x))dx + (\cos(x)+x\cos(y))dy = 0 \)
Problem Statement
\( (\sin(y)  y\sin(x))dx + (\cos(x)+x\cos(y))dy = 0 \)
Solution
video by blackpenredpen 

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\( (4x^210y)dx + (2x^3/y15x)dy = 0, y(2) = 1 \)
Problem Statement 

\( (4x^210y)dx + (2x^3/y15x)dy = 0, y(2) = 1 \)
Hint 

This is almost exact. Use the special integrating factor \(\mu(x,y) = x^m y^n\)
Problem Statement
\( (4x^210y)dx + (2x^3/y15x)dy = 0, y(2) = 1 \)
Hint
This is almost exact. Use the special integrating factor \(\mu(x,y) = x^m y^n\)
Solution
video by blackpenredpen 

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\( dy/dx = \cos(x)  y\sec(x), y(0) = 2 \)
Problem Statement
\( dy/dx = \cos(x)  y\sec(x), y(0) = 2 \)
Solution
video by blackpenredpen 

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\( (xy + y^2 + x^2)dx  x^2dy = 0 \)
Problem Statement
\( (xy + y^2 + x^2)dx  x^2dy = 0 \)
Solution
video by blackpenredpen 

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\( (2x+\cos y)dx + (x^2 + \tan y + \cos^2 y)dy = 0 \)
Problem Statement
\( (2x+\cos y)dx + (x^2 + \tan y + \cos^2 y)dy = 0 \)
Solution
video by blackpenredpen 

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\(\displaystyle{ (x^2+1)\frac{dy}{dx} + xy  x = 0 }\)
Problem Statement
\(\displaystyle{ (x^2+1)\frac{dy}{dx} + xy  x = 0 }\)
Solution
video by blackpenredpen 

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\( dy/dx = 2\sqrt{ 2xy+3 } \)
Problem Statement
\( dy/dx = 2\sqrt{ 2xy+3 } \)
Solution
video by blackpenredpen 

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\( 2xye^{x^2} dx + (e^{x^2} 1/y)dy = 0 \)
Problem Statement
\( 2xye^{x^2} dx + (e^{x^2} 1/y)dy = 0 \)
Solution
video by blackpenredpen 

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\( e^y dx  e^x dy = 0, y(0) = 0 \)
Problem Statement
\( e^y dx  e^x dy = 0, y(0) = 0 \)
Solution
video by blackpenredpen 

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\( dy/dx = y(1+xy^2), y(0) = 1/2 \)
Problem Statement
\( dy/dx = y(1+xy^2), y(0) = 1/2 \)
Solution
video by blackpenredpen 

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\( (y^2x^2)dx  xydy = 0, y(1) = 3 \)
Problem Statement
\( (y^2x^2)dx  xydy = 0, y(1) = 3 \)
Solution
He works this several different ways. Make sure to watch the complete video so that you can learn how to solve this in more ways than one.
video by blackpenredpen 

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\( dP/dt = P(ab\ln P), P(0) = P_0 \)
Problem Statement
\( dP/dt = P(ab\ln P), P(0) = P_0 \)
Solution
video by blackpenredpen 

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\( dP/dt = kP(1P/M), P(0) = P_0 \)
Problem Statement
\( dP/dt = kP(1P/M), P(0) = P_0 \)
Solution
video by blackpenredpen 

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\(\displaystyle{ \frac{dy}{dx} = \frac{xy}{x+y} }\)
Problem Statement
Find the general solution to the differential equation \(\displaystyle{ \frac{dy}{dx} = \frac{xy}{x+y} }\)
Solution
Here are two videos by two different instructors with similar solutions to this differential equation.
video by SyberMath 

video by blackpenredpen 

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\( dy/dx= \sqrt{y}  y \)
Problem Statement
\( dy/dx= \sqrt{y}  y \)
Solution
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\(\displaystyle{ dy/dx = \frac{12y\sin^2(x^2y)}{x} }\)
Problem Statement
\(\displaystyle{ dy/dx = \frac{12y\sin^2(x^2y)}{x} }\)
Solution
video by blackpenredpen 

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\( (3y+ x^2y^2)dx + (x2x^3y+x^4/y)dy = 0 \)
Problem Statement
\( (3y+ x^2y^2)dx + (x2x^3y+x^4/y)dy = 0 \)
Solution
video by blackpenredpen 

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\( 2xye^{x^2}dx + (e^{x^2}1/y)dy = 0 \)
Problem Statement
\( 2xye^{x^2}dx + (e^{x^2}1/y)dy = 0 \)
Solution
video by blackpenredpen 

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\(\displaystyle{ y' = x + \frac{1}{2x}y, y(1) = 0 }\)
Problem Statement
\(\displaystyle{ y' = x + \frac{1}{2x}y, y(1) = 0 }\)
Solution
video by blackpenredpen 

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\( y' = x^2 + 2xy + y^2, y(1) = 0 \)
Problem Statement
\( y' = x^2 + 2xy + y^2, y(1) = 0 \)
Solution
video by blackpenredpen 

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Intermediate
\(\displaystyle{ dy/dx = \frac{xy^2\sin x \cos x}{y(1x^2)}, y(0) = 4 }\)
Problem Statement
\(\displaystyle{ dy/dx = \frac{xy^2\sin x \cos x}{y(1x^2)}, y(0) = 4 }\)
Solution
video by blackpenredpen 

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\(\displaystyle{ y' = x + \frac{1}{2x}y + y^2, y(1) = 0 }\)
Problem Statement 

\(\displaystyle{ y' = x + \frac{1}{2x}y + y^2, y(1) = 0 }\)
Hint 

Start with \( y_1 = x^n\) and find \( y = y_1 + v \)
Problem Statement
\(\displaystyle{ y' = x + \frac{1}{2x}y + y^2, y(1) = 0 }\)
Hint
Start with \( y_1 = x^n\) and find \( y = y_1 + v \)
Solution
video by blackpenredpen 

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\( y = \left( \dfrac{dy}{dx} \right)^2 \)
Problem Statement
Solve the differential equation \( y = \left( \dfrac{dy}{dx} \right)^2 \)
Solution
video by SyberMath 

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Advanced
\( y = xy' + \sqrt{(y')^2 + 1} \)
Problem Statement 

\( y = xy' + \sqrt{(y')^2 + 1} \)
Hint 

Start by taking the derivative of the original differential equation.
Problem Statement
\( y = xy' + \sqrt{(y')^2 + 1} \)
Hint
Start by taking the derivative of the original differential equation.
Solution
video by blackpenredpen 

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\( y=xy'e^{y'} \)
Problem Statement
\( y=xy'e^{y'} \)
Solution
video by blackpenredpen 

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Really UNDERSTAND Differential Equations
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all first order techniques 
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