Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

3.1E: Exercises

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)


    Exercise \(\PageIndex{1}\)

    Find the order of the equation.

    (a) \(\displaystyle{d^2y\over dx^2}+2{dy\over dx}\ {d^3y\over dx^3}+x=0\)

    (b) \(y''-3y'+2y=x^7\)

    (c) \(y'-y^7=0\)

    (d) \(y''y-(y')^2=2\)


    (a) y''' therefore 3

    (b) y'' therefore 2

    (c) y' therefore 1

    (d) y'' therefore 2

    Exercise \(\PageIndex{2}\)

    Verify that the function is a solution of the differential equation on some interval, for any choice of the arbitrary constants appearing in the function.

    (a) \(y=ce^{2x}; \quad y'=2y\)

    (b) \(y={x^2\over3}+{c\over x}; \quad xy'+y=x^2\)

    (c) \(y={1\over2}+ce^{-x^2}; \quad y'+2xy=x\)

    (d) \(y=(1+ce^{-x^2/2}); (1-ce^{-x^2/2})^{-1} \quad 2y'+x(y^2-1)=0\)

    (e) \(y={\tan\left( {x^3\over3}+c\right)}; \quad y'=x^2(1+y^2)\)

    (f) \(y=(c_1+c_2x)e^x+\sin x+x^2; \quad y''-2y'+y=-2 \cos x+x^2-4x+2\)

    (g) \(y=c_1e^x+c_2x+{2\over x}; \quad (1-x)y''+xy'- y=4(1-x-x^2)x^{-3}\)

    (h) \(y=x^{-1/2}(c_1\sin x+c_2 \cos x)+4x+8\);



    (a) Given \(y=ce^{2x} \) then \(y' = ce^{2x} * 2 = 2y\).

    (c) Given \(y={1\over2}+ce^{-x^2} \) then \(y' = ce^{-x^2}*-2x\).

    Substituting into \(y'+2xy=x\) we obtain \(ce^{-x^2}*-2x +2x({1\over2}+ce^{-x^2}) = x\).

    Rearranging and simplifying, we obtain \(-2xe^{-x^2}+2xe^{-x^2}+x=x\) and thus \(x=x\).

    (e) Given \(y=\tan\left( \frac{x^3}{3}+c \right) \) then \(y' = x^2 \sec^2 \left( \frac{x^3}{3}+c \right)\).

    Substituting into \( y'=x^2\left(1+y^2 \right) \) we obtain \( y' = x^2\left(1+ \tan^2\left( \frac{x^3}{3}+c \right) \right) \).

    Recall from trig that \( 1+\tan^2\left(x\right)=\sec^2\left(x\right) \), thus \(y'=x^2 \sec^2 \left( \frac{x^3}{3}+c \right)\).

    (g) Given \(y=c_1e^x+c_2x+{2\over x} \) then \(y'=c_1e^x+c_2-\frac{2}{x^2} \) and \(y''=c_1e^x+\frac{4}{x^3} \)

    Substituting into \( (1-x)y''+xy'- y \) we obtain \( (1-x)(c_1e^x+\frac{4}{x^3}) +x(c_1e^x+c_2-\frac{2}{x^2}) -(c_1e^x+c_2x+\frac{2}{x}) \)

    Simplifying we obtain \( (1-x)y''+xy'- y = \frac{4}{x^3}-\frac{4}{x^2}-\frac{4}{x} \).

    Exercise \(\PageIndex{3}\)

    Find all solutions of the equation.

    (a) \(y'=-x\)

    (b) \(y'=-x \sin x\)

    (c) \(y'=x \ln x\)

    (d) \(y''=x \cos x\)

    (e) \(y''=2xe^x\)

    (f) \(y''=2x+\sin x+e^x\)

    (g) \(y'''=-\cos x\)

    (h) \(y'''=-x^2+e^x\)

    (i) \(y'''=7e^{4x}\)


    (a) \(y=\frac{-x^2}{2}+c \)

    (b) \(y=x \cos(x) - \sin(x) + c \)

    (c) \(y=\frac{x^2 ln(x)}{2}-\frac{x^2}{4}+c \)

    (d) \(y'=x\sin\left(x\right)+\cos\left(x\right) + c_1\) thus \(y=2\sin\left(x\right)-x\cos\left(x\right)+c_1x +c_2\)

    (e) \(y'=2(x-1)e^x+c_1 \) thus \(y = 2(x-2)e^x+c_1x+c_2 \)

    (f) \(y'=x^2-\cos\left(x\right)+e^x+c_1 \) thus \(y=\frac{x^3}{3}-\sin\left(x\right)+e^x+c_1x+c_2 \)

    (g) \(y''=-sin\left(x\right)+c_1\) thus \(y'=\cos\left(x\right)+c_1x+c_2\) resulting in \(y=\sin\left(x\right)+\frac{c_1x^2}{2}+c_2x+c_3\).

    (h)\(y''=\frac{-x^3}{3} + e^x +c_1\) thus \(y'=\frac{-x^4}{12}+e^x+c_1x+c_2\) resulting in \(y=\frac{-x^5}{60}+e^x+c_1x^2+c_2x+c_3\).

    (i) \(y''=\frac{7e^{4x}}{4}+c_1\) thus \(y'=\frac{7e^{4x}}{16}+c_1x+c_2\) resulting in \(y=\frac{7e^{4x}}{64}+c_1x^2+c_2x+c_3\).

    Exercise \(\PageIndex{4}\)

    Solve the initial value problem.

    (a) \(y'=-xe^x, \quad y(0)=1\)

    (b) \(y'=x \sin x^2, \quad y\left({\sqrt{\pi\over2}}\right)=1\)

    (c) \(y'=\tan x, \quad y(\pi/4)=3\)

    (d) \(y''=x^4, \quad y(2)=-1, \quad y'(2)=-1\)

    (e) \(y''=xe^{2x}, \quad y(0)=7, \quad y'(0)=1\)

    (f) \(y''=- x \sin x, \quad y(0)=1, \quad y'(0)=-3\)

    (g) \(y'''=x^2e^x, \quad y(0)=1, \quad y'(0)=-2, \quad y''(0)=3\)

    (h) \(y'''=2+\sin 2x, \quad y(0)=1, \quad y'(0)=-6, \quad y''(0)=3\)

    (i) \(y'''=2x+1, \quad y(2)=1, \quad y'(2)=-4, \quad y''(2)=7\)


    (a) \(y=-xe^x+e^x+c\). Since \(y(0)=1, c=0\). Thus the solution is \(y=-xe^x+e^x\).

    (c) \(y=-\ln\left(\left|\cos\left(x\right)\right|\right)+c\). Since \(y(\frac{\pi}{4})=3\), \(c=3+\ln(\frac{\sqrt{2}}{2}) \). Thus the solution is \(y=-\ln\left(\left|\cos\left(x\right)\right|\right)+3+\ln(\frac{\sqrt{2}}{2}) \).

    (e) \(y'=\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}+c\). Since \(y'(0)=1, c=\frac{5}{4}\).

    \(y= \frac{xe^{2x}}{4}-\frac{e^{2x}}{4}+\frac{5x}{4}+c\). Since \(y(0)=7, c=\frac{29}{4}\).

    Thus the solution is \(y=\frac{xe^{2x}}{4}-\frac{e^{2x}}{4}+\frac{5x}{4}+\frac{29}{4}\)

    (f) \( y'=-x\cos\left(x\right)-\sin\left(x\right)+c_1\). Since \(y'(0)=-3, c_1=-3\).

    \(y= x\sin\left(x\right)+2\cos\left(x\right)-3x+c_2\). Since \(y(0)=1, c_2=-1\).

    Thus the solution is \(y=x\sin\left(x\right)+2\cos\left(x\right)-3x-1\).

    (h) \( y''=2x-\frac{\cos\left( 2x\right)}{2}+c \). Since \(y''(0)=3\), \(c=\frac{7}{2}\).

    \(y'=x^2-\frac{\sin\left( 2x\right)}{4}+\frac{7x}{2}+c \). Since \(y'(0)=-6\), \(c=-6\).

    \(y=\frac{x^3}{3}+\frac{\cos\left( 2x\right)}{8}+\frac{7x^2}{4}-6x+c\). Since \(y(0)=1\), \(c=\frac{7}{8}\).

    Thus the solution is \(y=\frac{x^3}{3}+\frac{\cos\left( 2x\right)}{8}+\frac{7x^2}{4}-6x+\frac{7}{8}\).

    Exercise \(\PageIndex{5}\)

    Verify that the function is a solution of the initial value problem.

    (a) \(y=x\cos x; \quad y'=\cos x-y\tan x, \quad y(\pi/4)={\pi\over4\sqrt{2}}\)

    (b) \({y={1+2\ln x\over x^2}+{1\over2}; \quad y'={x^2-2x^2y+2\over x^3}, \quad y(1)={3\over2}}\)

    (c) \(y={\tan\left({x^2\over2}\right)}; \quad y'=x(1+y^2), \quad y(0)=0\)

    (d) \({y={2\over x-2}; \quad y'={-y(y+1)\over x}}, \quad y(1)=-2\)


    (a) Substituting for \(y\) in \(y'\) we obtain: \(y'=\cos(x)-x\sin(x)\). Integrating we obtain \(y=x\cos(x) + C\). Given that \( y(\frac{\pi}{4})=\frac{4\pi}{\sqrt{2}}\), then \(C=0\) and thus \(y=x\cos(x)\).

    (c) Substituting for \(y\) in \(y'\) we obtain: \(y'=x+x\tan^2(\frac{x^2}{2}) \). Integrating we obtain \( y=\tan(\frac{x^2}{2})+C\). Given that \( y(0)=0 \), then \( C=0 \) and thus \( y=\tan(\frac{x^2}{2})\).

    Exercise \(\PageIndex{6}\)

    Verify that the function is a solution of the initial value problem.

    (a) \(y=x^2(1+\ln x); \quad y''={3xy'-4y\over x^2}, \quad y(e)=2e^2, \quad y'(e)=5e\)

    (b) \(y={x^2\over3}+x-1; \quad y''={x^2-xy'+y+1\over x^2}, \quad y(1)={1\over3}, \quad y'(1)={5\over3}\)

    (c) \(y=(1+x^2)^{-1/2}; \quad y''={(x^2-1)y-x(x^2+1)y'\over (x^2+1)^2}, \quad y(0)=1, y'(0)=0\)

    (d) \(y={x^2\over 1-x}; \quad y''={2(x+y)(xy'-y)\over x^3}, \quad y(1/2)=1/2, \quad y'(1/2)=3\)


    (a) Given \(y=x^2(1+\ln x)\), then \(y'=3x+2x\ln|x|\). Substituting \(y\) and \(y'\) into \(y''\) we obtain \(y''=\frac{3x(3x+2x\ln|x|)-4(x^2(1+\ln|x|)}{x^2}\). Simplifying we obtain \(y''=5+2\ln|x|\) and then integrating, we obtain \( y'= 5x+2(x\ln|x|-x)+ c\). Since \(y'(e)=5e\), \(c=0\). Integrating \( y'= 3x+2x\ln|x|\), we obtain \(y = \frac{3x^2}{2} + x^2\ln|x|-\frac{x^2}{2}+c = x^2(1+\ln|x|)+c\). Since \(y(e) = 2e^2\), \(c= 0 \).

    (c) Given \(y=(1+x^2)^{-1/2}\), then \(y'=\frac{-x}{(x^2+1)^{\frac{3}{2}}} \). Substituting for \(y\) and \(y'\) into \(y''\), and simplifying, we obtain \(y''=\frac{2x^2-1}{(1+x^2)^{5/2}} \). Integrating we obtain \(y'=\frac{-x}{(x^2+1)^{3/2}}+c\). Since \(y'(0)=0\), \(c=0\). Integrating \( \frac{-x}{(x^2+1)^{3/2}} \), we obtain \(y=\frac{1}{(x^2+1)^{1/2}} +c \). Since \(y(0)=1\), \(c=0\).

    Exercise \(\PageIndex{7}\)

    Suppose an object is launched from a point 320 feet above the earth with an initial velocity of 128 ft/sec upward, and the only force acting on it thereafter is gravity. Take \(g=32 ft/sec^2\)

    (a) Find the highest altitude attained by the object.

    (b) Determine how long it takes for the object to fall to the ground.


    (a) Since \(y'' =-32\), \(y'=-32t+c\). Since \(y'(0)=128\), \(c= 128\). Similarly since \(y'=-32t+128\) and \(y(0)=320\), \(y=-16t^2+128t+320\). The object is at its maximum height when \(y'=0\). This occurs at \(t=4\), which corresponds to a height of 576 feet.

    (b) Set \(y=0\) and solvng for \(t\), we obtain \(t=10s\).

    Exercise \(\PageIndex{8}\)

    Let \(a\) be a nonzero real number.

    (a) Verify that if \(c\) is an arbitrary constant then equation A: \(y=(x-c)^a \) is a solution of equation B: \(y'=ay^{(a-1)/a}\) on \((c,\infty)\).

    (b) Suppose \(a<0\) or \(a>1\). Can you think of a solution of (B) that isn't of the form (A)?


    Add texts here. Do not delete this text first.

    Exercise \(\PageIndex{9}\)

    Verify that \(y= e^x-1, x \ge 0\) and \(1-e^{-x}, x < 0, \) is a solution of \(y'=|y|+1\) on \((-\infty,\infty)\).


    Use the definition of derivative at \(x=0\)


    Add texts here. Do not delete this text first.

    Exercise \(\PageIndex{10}\)

    (a) Verify that if \(c\) is any real number then equation A: \(y=c^2+cx+2c+1\) satisfies equation B: \(y'={-(x+2)+\sqrt{x^2+4x+4y}\over2}\) on some open interval. Identify the open interval.

    (b) Verify that \(y_1={-x(x+4)\over4}\) also satisfies (B) on some open interval, and identify the open interval. (Note that \(y_1\) can't be obtained by selecting a value of \(c\) in (A).


    Add texts here. Do not delete this text first

    This page titled 3.1E: Exercises is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by OpenStax.

    • Was this article helpful?