Answers to Exercises for Chapter 2

    1. $$y = c x^3 $$
    2. $$y=c e^{-x^2} + (x^2-1)/ 2 $$ (Use integrating factor, then integration by parts)
    3. $$x = c e^{ y^2+3t} -2 $$
    4. $$ x = c e^{-t} + e^t/2 $$
    5. $$y = - \frac{1}{x^2(x+c)}$$ (set \(y=1/u \) to get a first-order linear equation for \(u\), see p. 18)

    1. $$y=\frac{1}{\sqrt[3]{1+x^3}} $$
    2. $$y =\frac{5-\cos(2x)}{4\cos x} $$
    3. $$y = \frac{x(2x^3+3x^2+1)}{1+x}$$ (use partial fractions to find integrating factor)
    4. $$ x =\tan((t^2-1)/2)$$
  3. Suppose that \(y' = f(y/x) \). Set \( u(x) = y(x)/x\), so \(y=x u, \ y' = u+ x u'\). Then \(u \) obeys the equation $$ u+xu' = f(u)$$ Hence $$u' =\frac{f(u)-u}{x},$$ which is separable. For $$ y' = \frac{ x^2 + y^2}{xy}= \frac{1+y^2/x^2}{y/x}, $$ $$x u' =\frac{1 + u^2 }{u}-u= \frac{1}{u}$$ which leads to $$y= \pm x \sqrt{2\ln x + c}$$ For the initial condition \(y(1)=1\), \(c=1\) and we take the + sign.

    1. $$ x = a e^{-3t/2} + be^{2y/3}$$
    2. $$ x = a e^{-3t/2} + be^{2y/3} + e^{-2t} -2 $$
    3. $$ y = a x^2 + b / x^2 $$ (try \(y=ax^p \), see Example 2.14)
    4. $$y = a x^2+ b/x^2 -3x $$ (similar to particular integral method)

    1. $$ x=e^{-3t}$$
    2. $$y = e^{3x-3}$$
    3. $$ y=(\cos(3x)+6\cos(4x))/7 $$

  6. See Remark 2.2. Auxiliary equations is $$m^3+ m^2 + m + 1 = (m+1)(m^2+1)=0$$ with solutions \(m=-1, \ \pm i\). Hence $$y = ae^{-x} + b\cos x + c \sin x . $$ $$y'''+y''+y'+y=0. $$ If \(y\) is bounded and \(y(0)=0\) there can be no exponential term and no cosine term so \( y=\sin x\). If the RHS of the equation is changed from 0 to \( \sin(2x)\), the general solution has an extra term $$ (2 \cos 2x - \sin 2x)/15$$ and the specific solution is $$ y= (2 \cos 2x -2\cos x - \sin 2x +17\sin x)/15$$

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