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# 3.1E: Exercises

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

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

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

Answer

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

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

Answer

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

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

Answer

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

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

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

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

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

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

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

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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

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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

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