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Mathematics LibreTexts

2.1E: Linear First Order Equations (Exercises)

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    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 \(\lim_{t\to\infty}G(t)\).


    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


    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}



    [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




    respectively, and \(c_1\) and \(c_2\) are constants, then \(y=c_1y_1+c_2y_2\) is a solution of


    (This is theprinciple of superposition.)



    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

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



    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


    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


    [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) \(

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    [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

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