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# 2.1E: Linear First Order Equations (Exercises)

• • Contributed by William F. Trench
• Andrew G. Cowles Distinguished Professor Emeritus (Mathamatics) at Trinity University

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#### Exercises

In Exercises [exer:2.1.1}[exer:2.1.5} find the general solution.

[exer:2.1.1] $$y'+ay=0$$ ( $$a$$=constant) [exer:2.1.2] $$y'+3x^2y=0$$ [exer:2.1.3] $$xy'+(\ln x)y=0$$ [exer:2.1.4] $$xy'+3y=0$$ [exer:2.1.5] $$x^2y'+y=0$$

[exer:2.1.6] $${y'+\left({1+x\over x}\right)y=0,\quad y(1)=1}$$

[exer:2.1.7] $${xy'+\left(1+{1\over\ln x}\right)y=0,\quad y(e)=1}$$

[exer:2.1.8] $${xy'+(1+ x\cot x)y=0,\quad y\left({\pi\over 2} \right)=2}$$

[exer:2.1.9] $${y'-\left({2x\over 1+x^2}\right)y=0,\quad y(0)=2}$$

[exer:2.1.10] $${y'+{k\over x}y=0,\quad y(1)=3 \quad\mbox{($$k $$= constant)}}$$

[exer:2.1.11] $$y'+(\tan kx)y=0,\quad y(0)=2 \quad\mbox{($$k= $$constant)}$$

[exer:2.1.12] $$y'+3y=1$$ [exer:2.1.13] $${y'+\left({1\over x}- 1\right)y=-{2\over x}}$$ [exer:2.1.14] $$y'+2xy=xe^{-x^2}$$ [exer:2.1.15] $${y'+{2x\over1+x^2}y={e^{-x}\over1+x^2}}$$

[exer:2.1.16] $${y'+{1\over x}y={7\over x^2}+3}$$ [exer:2.1.17] $${y'+{4\over x-1}y = {1\over (x-1)^5}+{\sin x\over (x-1)^4}}$$ [exer:2.1.18] $$xy'+(1+2x^2)y=x^3e^{-x^2}$$

[exer:2.1.19] $${xy'+2y={2\over x^2}+1}$$ [exer:2.1.20] $$y'+(\tan x)y=\cos x$$ [exer:2.1.21] $${(1+x)y'+2y={\sin x \over 1 + x}}$$

[exer:2.1.22] $$(x-2)(x-1)y'-(4x-3)y=(x-2)^3$$

[exer:2.1.23] $$y'+(2\sin x\cos x) y=e^{-\sin^2x}$$ [exer:2.1.24] $$x^2y'+3xy=e^x$$

[exer:2.1.25] $$y'+7y=e^{3x},\quad y(0)=0$$

[exer:2.1.26] $${(1+x^2)y'+4xy={2\over 1+x^2},\quad y(0)=1}$$

[exer:2.1.27] $${xy'+3y={2\over x(1+x^2)},\quad y(-1)=0}$$

[exer:2.1.28] $${y'+ (\cot x)y=\cos x,\quad y\left({\pi\over 2}\right)=1}$$

[exer:2.1.29] $${y'+{1\over x}y={2\over x^2}+1,\quad y(-1)=0}$$

[exer:2.1.30] $${(x-1)y'+3y={1\over (x-1)^3} + {\sin x\over (x-1)^2},\quad y(0)=1}$$

[exer:2.1.31] $$xy'+2y=8x^2,\quad y(1)=3$$

[exer:2.1.32] $$xy'-2y=-x^2,\quad y(1)=1$$

[exer:2.1.33] $$y'+2xy=x,\quad y(0)=3$$

[exer:2.1.34] $${(x-1)y'+3y={1+(x-1)\sec^2x\over (x-1)^3},\quad y(0)=-1}$$

[exer:2.1.35] $${(x+2)y'+4y={1+2x^2\over x(x+2)^3},\quad y(-1)=2}$$

[exer:2.1.36] $$(x^2-1)y'-2xy=x(x^2-1),\quad y(0)=4$$

[exer:2.1.37] $$(x^2-5)y'-2xy=-2x(x^2-5),\quad y(2)=7$$

[exer:2.1.38] $$y'+2xy=x^2,\quad y(0)=3$$

[exer:2.1.39] $${y'+{1\over x}y={\sin x\over x^2},\quad y(1)=2}$$

[exer:2.1.40] $${y'+y={e^{-x}\tan x\over x},\quad y(1)=0}$$

[exer:2.1.41] $${y'+{2x\over 1+x^2}y={e^x\over (1+x^2)^2}, \quad y(0)=1}$$

[exer:2.1.42] $$xy'+(x+1)y=e^{x^2},\quad y(1)=2$$

[exer:2.1.43] 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 $$\lim_{t\to\infty}G(t)$$.

[exer:2.1.44]

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

Show that the general solution of (A) on $$(-\infty,0)$$ and $$(0,\infty)$$ is

$y={1\over2}+cx^2.$

Show that $$y$$ is a solution of (A) on $$(-\infty,\infty)$$ if and only if $y=\left\{\begin{array}{ll} {{1\over2}+c_1x^2}, &x \ge 0,\\[4pt] {{1\over2}+c_2x^2}, &x < 0,\end{array}\right.$ where $$c_1$$ and $$c_2$$ are arbitrary constants.

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

Use 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 does’nt contradict Theorem [thmtype:2.1.1}

## b

.

[exer:2.1.45] Suppose $$f$$ is continuous on an open interval $$(a,b)$$ and $$\alpha$$ is a constant.

Derive a formula for the solution of the initial value problem

$y'+\alpha y=f(x),\quad y(x_0)=y_0, \eqno{\rm (A)}$

where $$x_0$$ is in $$(a,b)$$ and $$y_0$$ is an arbitrary real number.

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

[exer:2.1.46] Assume that all functions in this exercise are defined on a common interval $$(a,b)$$.

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 theprinciple of superposition.)

Use

## a

to show that if $$y_1$$ and $$y_2$$ are solutions of the nonhomogeneous equation

$y'+p(x)y=f(x), \eqno{\rm (A)}$

then $$y_1-y_2$$ is a solution of the homogeneous equation

$y'+p(x)y=0. \eqno{\rm (B)}$

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).

[exer:2.1.47] 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

$z'+p(x)z=f(x).$

[exer:2.1.48] Solve by the method discussed in Exercise [exer:2.1.47}.

(a) $$(\sec^2y)y'- 3\tan y=-1$$ (b) $${e^{y^2}\left(2yy'+ {2\over x}\right) ={1\over x^2}}$$ (c) $$ParseError: EOF expected (click for details) Callstack: at (Bookshelves/Differential_Equations/Book:_Elementary_Differential_Equations_with_Boundary_Values_Problems_(Trench)/2:_First_Order_Equations/2.1:_Linear_First_Order_Equations/2.1E:_Linear_First_Order_Equations_(Exercises)), /content/body/div/p/span, line 1, column 3 $$

[exer:2.1.49] We’ve shown that if $$p$$ and $$f$$ are continuous on $$(a,b)$$ then every solution of

$y'+p(x)y=f(x) \eqno{\rm(A)}$

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 $$f$$, $$f'$$, …, $$f^{(m)}$$ and $$p$$, $$p'$$, …, $$p^{(m-1)}$$ are continuous on $$(a,b)$$, where $$m$$ is a positive integer, and define

\begin{aligned} f_0&=&f,\\ f_j&=&f_{j-1}'+pf_{j-1},\quad 1\le j\le m.\end{aligned}

Show that

$u^{(j+1)}={f_j\over y_1},\quad 0\le j\le m.$