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# 3.2E: Exercises

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## Exercises

### Exercise $$\PageIndex{1}$$

Find the general solution.

(a) $$y'+ay=0$$ ($$a$$=constant)

(b) $$y'+3x^2y=0$$

(c) $$xy'+(\ln x)y=0$$

(d) $$xy'+3y=0$$

(e) $$x^2y'+y=0$$

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### Exercise $$\PageIndex{2}$$

Solve the initial value problem.

(a) $${y'+\left({1+x\over x}\right)y=0,\quad y(1)=1}$$

(b) $${xy'+\left(1+{1\over\ln x}\right)y=0,\quad y(e)=1}$$

(c) $${xy'+(1+ x\cot x)y=0,\quad y\left({\pi\over 2} \right)=2}$$

(d) $${y'-\left({2x\over 1+x^2}\right)y=0,\quad y(0)=2}$$

(e) $${y'+{k\over x}y=0,\quad y(1)=3 \quad {(k=constant)}}$$

(f) $$y'+(\tan kx)y=0,\quad y(0)=2 \quad {(k=constant)}$$

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### Exercise $$\PageIndex{3}$$

Find the general solution. Also, plot a direction field and some integral curves on the rectangular region $$\{-2\le x\le2,\ -2\le y\le2$$\}.}

(a) $$y'+3y=1$$

(b) $${y'+\left({1\over x}-1\right)y=-{2\over x}}$$

(c) $$y'+2xy=xe^{-x^2}$$

(d) $${y'+{2x\over1+x^2}y={e^{-x}\over1+x^2}}$$

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### Exercise $$\PageIndex{4}$$

Find the general solution.

(a) $${y'+{1\over x}y={7\over x^2}+3}$$

(b) $${y'+{4\over x-1}y = {1\over (x-1)^5}+{\sin x\over (x-1)^4}}$$

(c) $$xy'+(1+2x^2)y=x^3e^{-x^2}$$

(d) $${xy'+2y={2\over x^2}+1}$$

(e) $$y'+(\tan x)y=\cos x$$

(f) $${(1+x)y'+2y={\sin x \over 1 + x}}$$

(g) $$(x-2)(x-1)y'-(4x-3)y=(x-2)^3$$

(h) $$y'+(2\sin x\cos x) y=e^{-\sin^2x}$$

(i) $$x^2y'+3xy=e^x$$

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### Exercise $$\PageIndex{5}$$

Solve the initial value problem and sketch the graph of the solution.

(a) $$y'+7y=e^{3x},\quad y(0)=0$$

(b) $${(1+x^2)y'+4xy={2\over 1+x^2},\quad y(0)=1}$$

(c) $${xy'+3y={2\over x(1+x^2)},\quad y(-1)=0}$$

(d) $${y'+ (\cot x)y=\cos x,\quad y\left({\pi\over 2}\right)=1}$$

(e) $${y'+{1\over x}y={2\over x^2}+1,\quad y(-1)=0}$$

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### Exercise $$\PageIndex{6}$$

Solve the initial value problem.

(a) $${(x-1)y'+3y={1\over (x-1)^3} + {\sin x\over (x-1)^2},\quad y(0)=1}$$

(b) $$xy'+2y=8x^2,\quad y(1)=3$$

(c) $$xy'-2y=-x^2,\quad y(1)=1$$

(d) $$y'+2xy=x,\quad y(0)=3$$

(e) $${(x-1)y'+3y={1+(x-1)\sec^2x\over (x-1)^3},\quad y(0)=-1}$$

(f) $${(x+2)y'+4y={1+2x^2\over x(x+2)^3},\quad y(-1)=2}$$

(g) $$(x^2-1)y'-2xy=x(x^2-1),\quad y(0)=4$$

(h) $$(x^2-5)y'-2xy=-2x(x^2-5),\quad y(2)=7$$

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### Exercise $$\PageIndex{7}$$

Solve the initial value problem and leave the answer in a form involving a definite integral.

You can solve these problems numerically by methods discussed in Chapter~3.

(a) $$y'+2xy=x^2,\quad y(0)=3$$

(b) $${y'+{1\over x}y={\sin x\over x^2},\quad y(1)=2}$$

(c) $${y'+y={e^{-x}\tan x\over x},\quad y(1)=0}$$

(d) $${y'+{2x\over 1+x^2}y={e^x\over (1+x^2)^2}, \quad y(0)=1}$$

(e) $$xy'+(x+1)y=e^{x^2},\quad y(1)=2$$

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### Exercise $$\PageIndex{8}$$

Experiments indicate that glucose is absorbed by the body at a rate proportional to the amount of glucose present in the bloodstream. Let $$\lambda$$ denote the (positive) constant of proportionality. Now suppose glucose is injected into a patient's bloodstream at a constant rate of $$r$$ units per unit of time. Let $$G=G(t)$$ be the number of units in the patient's bloodstream at time $$t>0$$. Then $$G'=-\lambda G+r,$$ where the first term on the right is due to the absorption of the glucose by the patient's body and the second term is due to the injection. Determine $$G$$ for $$t>0$$, given that $$G(0)=G_0$$. Also, find $$\lim_{t\to\infty}G(t)$$.

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### Exercise $$\PageIndex{9}$$

(a) Plot a direction field and some integral curves for equation A: $$xy'-2y=-1$$ on the rectangular region $$\{-1\le x\le 1, -.5\le y\le 1.5\}$$. What do all the integral curves have in common?

(b) Show that the general solution of (A) on $$(-\infty,0)$$ and $$(0,\infty)$$ is $$y={1\over2}+cx^2.$$

(c) Show that $$y$$ is a solution of (A) on $$(-\infty,\infty)$$ if and only if

$$y=1 \over 2+c_1x^2, x\ge 0$$ and $$1\over 2+c_2x^2, x < 0,$$.

where $$c_1$$ and $$c_2$$ are arbitrary constants.

(d) Conclude from \part{c} that all solutions of (A) on $$(-\infty,\infty)$$ are solutions of the initial value problem $$xy'-2y=-1,\quad y(0)={1\over2}.$$

(e) Use part b to show that if $$x_0\ne0$$ and $$y_0$$ is arbitrary, then the initial value problem $$xy'-2y=-1,\quad y(x_0)=y_0$$ has infinitely many solutions on $$(-\infty,\infty)$$. Explain why this doesn't contradict Theorem~\ref{thmtype:3.3.1} \part{b}.

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### Exercise $$\PageIndex{10}$$

Suppose $$f$$ is continuous on an open interval $$(a,b)$$ and $$\alpha$$ is a constant.

(a) Derive a formula for the solution of the initial value problem $$y'+\alpha y=f(x),\quad y(x_0)=y_0,$$ where $$x_0$$ is in $$(a,b)$$ and $$y_0$$ is an arbitrary real number.

(b) Suppose $$(a,b)=(a,\infty)$$, $$\alpha > 0$$ and $$\displaystyle{\lim_{x\to\infty} f(x)=L}$$. Show that if $$y$$ is the solution of (a), then

$$\displaystyle{\lim_{x\to \infty} y(x)=L/\alpha}$$.

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### Exercise $$\PageIndex{11}$$

Assume that all functions in this exercise are defined on a common interval $$(a,b)$$

(a) Prove: If $$y_1$$ and $$y_2$$ are solutions of $$y'+p(x)y=f_1(x)$$ and $$y'+p(x)y=f_2(x)$$ respectively, and $$c_1$$ and $$c_2$$ are constants, then $$y=c_1y_1+c_2y_2$$ is a solution of $$y'+p(x)y=c_1f_1(x)+c_2f_2(x).$$

(This is the principle of superposition)

(b) Use (a) to show that if $$y_1$$ and $$y_2$$ are solutions of the nonhomogeneous equation $$y'+p(x)y=f(x),$$ then $$y_1-y_2$$ is a solution of the homogeneous equation B: $$y'+p(x)y=0$$.

(c) Use (a) to show that if $$y_1$$ is a solution of (A) and $$y_2$$ is a solution of (B), then $$y_1+y_2$$ is a solution of (A).

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### Exercise $$\PageIndex{12}$$

Some nonlinear equations can be transformed into linear equations by changing the dependent variable. Show that if $$g'(y)y'+p(x)g(y)=f(x)$$ where $$y$$ is a function of $$x$$ and $$g$$ is a function of $$y$$, then the new dependent variable $$z=g(y)$$ satisfies the linear equation $$z'+p(x)z=f(x).$$

### Exercise $$\PageIndex{13}$$
We've shown that if $$p$$ and $$f$$ are continuous on $$(a,b)$$ then every solution of equation A: $$y'+p(x)y=f(x)$$ on $$(a,b)$$ can be written as $$y=uy_1$$, where $$y_1$$ is a nontrivial solution of the complementary equation for (A) and $$u'=f/y_1$$. Now suppose $$f$$, $$f'$$, \dots, $$f^{(m)}$$ and $$p$$, $$p'$$, \dots, $$p^{(m-1)}$$ are continuous on $$(a,b)$$, where $$m$$ is a positive integer, and define \begin{eqnarray*} f_0&=&f,\\ f_j&=&f_{j-1}'+pf_{j-1},\quad 1\le j\le m. \end{eqnarray*}
Show that $$u^{(j+1)}={f_j\over y_1},\quad 0\le j\le m.$$