Opening the book…
In the following exercises, determine the order of each differential equation.
1) \(y′+y=3y^2\)
- Answer
- 1st-order
2) \((y′)^2=y′+2y\)
- Answer
- 1st-order
3) \(y'''+y''y′=3x^2\)
- Answer
- 3rd-order
4) \(y′=y''+3t^2\)
- Answer
- 2nd-order
5) \(\dfrac{dy}{dx}+\dfrac{d^2y}{dx^2}=3x^4\)
- Answer
- 2nd-order
6) \(\left(\dfrac{dy}{dt}\right)^2+8\dfrac{dy}{dt}+3y=4t\)
- Answer
- 1st-order
In exercises 7-12, verify that the given function is a solution to the given differential equation.
7) \(y=2e^{−x}+x−1\quad\) solves \(\quad y′=x−y\)
8) \(y=e^{3x}−\dfrac{e^x}{2}\quad\) solves \(\quad y′=3y+e^x\)
9) \(y=2+x^{-3/2}\quad\) solves \(\quad 4x^2y''+12xy'+3y=6\)
10) \(y=e^x+\dfrac{\sin x}{2}−\dfrac{\cos x}{2}\quad\) solves \(\quad y′=\cos x+y\)
11) \(x^2y+y=4\quad\) solves \(\displaystyle \quad y′=-\frac{2xy}{x^2+2y}\)
- Hint
- Use implicit differentiation to find \(dy/dx\).
12) \(y=\sin(xy)\quad\) solves \(\displaystyle \quad y′=\frac{y \cos(xy)}{1-x\cos(xy)}\)
In the following exercises, verify the given general solution and find the particular solution.
13) Find the particular solution to the differential equation \(y′=3x^3\) that passes through \((1,4.75)\), given that \(y=C+\dfrac{3x^4}{4}\) is a general solution.
- Answer
- \(y=4+\dfrac{3x^4}{4}\)
14) Find the particular solution to the differential equation \(y′=2xy\) that passes through \(\left(0,\frac{1}{2}\right)\), given that \(y=Ce^{x^2}\) is a general solution.
- Answer
- \(y=\frac{1}{2}e^{x^2}\)
15) Find the particular solution to the differential equation \(y′x^2=y\) that passes through \(\left(1,\frac{2}{e}\right)\), given that \(y=Ce^{−1/x}\) is a general solution.
- Answer
- \(y=2e^{−1/x}\)
16) Find the particular solution to the differential equation \(\dfrac{du}{dt}=\tan u\) that passes through \(\left(1,\frac{π}{2}\right)\), given that \(u=\sin^{−1}\big(e^{C+t}\big)\) is a general solution.
- Answer
- \(u=\sin^{−1}\big(e^{−1+t}\big)\)
17) Find the particular solution to the differential equation \(y′(1−x^2)=1+y\) that passes through \((0,−2),\) given that \(y=C\dfrac{\sqrt{x+1}}{\sqrt{1−x}}−1\) is a general solution.
- Answer
- \(y=−\dfrac{\sqrt{x+1}}{\sqrt{1−x}}−1\)
In the following exercises, find the general solution to the differential equation.
18) \(y′=\ln x+\tan x\)
- Answer
- \(y=C−x+x\ln x−\ln(\cos x)\)
19) \(y′=4^x +e^x - 2x\)
- Answer
- \(y=C+\dfrac{4^x}{\ln 4} + e^x + x^2\)
20) \(y′=\sin^{−1}(2x)\)
- Answer
- \(\displaystyle y=x \arcsin \! \left(2 x \right)+\frac{\sqrt{1-4 x^{2}}}{2} +C \)
21) \(y′=2t\sqrt{t^2+16}\)
- Answer
- \(y=\frac{2}{3}\sqrt{t^2+16}\big(t^2+16\big)+C\)
22) \(x′=t\sqrt{4+t}\)
- Answer
- \(x=\frac{2}{15}\sqrt{4+t}\big(3t^2+4t−32\big)+C\)
23) \(y′=\dfrac{y}{x}\)
- Answer
- \(y=Cx\)
In the following exercises, solve the initial-value problems. Separate variables and integrate both sides.
24) \(\dfrac{dy}{dt}=−t; \; y(0)=1. \)
- Answer
- \(y=1−\dfrac{t^2}{2},\)
25) \(\dfrac{dy}{dt}=−y; \; y(0)=-1.\)
- Answer
- \(y=−e^{−t}\)
26) \(\dfrac{dy}{dx}=2xy+y; \; y(0)=3\)
- Answer
- Hint: the right-hand side is \(y(2x+1)\). Answer is \(y=3e^{x^2+x}\).
27) \(\dfrac{dy}{dx}= \dfrac{1+y^2}{\cos x}; \; y(0)=1\)
- Answer
- \(\displaystyle y=\tan \big( \ln |\sec x + \tan x| + \frac{\pi}{4} \big) \)
28) \(\dfrac{dy}{dx}= \dfrac{y^2}{x^2+x}; \; y(1)=\dfrac{1}{2\ln 2}\)
- Answer
- \(\displaystyle y=\frac{1}{\ln 2 + \ln|x+1|-\ln|x|} \)
This page titled 9.1E: Exercises for Section 9.1 was last modified on Wed, 16 Jul 2025 02:30:57 GMT and is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Gilbert Strang & Edwin “Jed” Herman (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform.
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