3.1E: Exercises

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$trench-1995.jpg$
Contributed by William F. Trench
Andrew G. Cowles Distinguished Professor Emeritus (Mathamatics) at Trinity University

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Exercises

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) $y''-3y'+2y=x^7$

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

Answer

(a) y''' therefore 3

(b) y'' therefore 2

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

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

$x^2y''+xy'+{\left(x^2-{1\over4}\right)}y=4x^3+8x^2+3x-2$

Answer

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

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) $y'=-x \sin x$

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

Answer

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

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

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

Answer

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

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

Answer

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

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

Answer

(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.

Answer

(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)?

Answer: 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 $(-\infty,\infty)$.

Hint: Use the definition of derivative at $x=0$

Answer: 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).

Answer: Add texts here. Do not delete this text first