2.4 HW 4

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2.4.1 problems description

2.4.2 Problem 1

part a

\begin{align*} y^{\prime } & =-\cos x+2ax+b\\ y^{\prime \prime } & =\sin x+2a\\ y^{\prime \prime \prime } & =\cos x \end{align*}

Substituting into the ODE \(y^{\prime \prime \prime }=\cos x\) shows it satsifies it. Hence this is true for any \(a,b,c\).

part b

Since \(\tan \left ( x+c\right ) =\frac{\sin \left ( x+c\right ) }{\cos \left ( x+c\right ) }\) then

\[ y^{\prime }=1+\tan ^{2}\left ( x+c\right ) \]

Substituting this into the ode \(y^{\prime }=1+y^{2}\) gives

\[ 1+\tan ^{2}\left ( x+c\right ) =1+\tan ^{2}\left ( x+c\right ) \]

Which is true for any \(c\)

2.4.3 Problem 2

see Key.

2.4.4 Problem 3

see Key

2.4.5 Problem 4

(a) Find all solutions to \(yy^{\prime }+25x=0\) (b) \(y^{\prime }=ky^{2}\) (c) \(xy^{\prime }=x+y\)

Part a

\begin{align*} y\frac{dy}{dx} & =-25x\\ ydy & =-25xdx\\ \frac{y^{2}}{2} & =-\frac{25}{2}x^{2}+C\\ y^{2} & =-25x^{2}+C_{1} \end{align*}

Hence

\[ y=\pm \sqrt{C_{1}-25x^{2}}\]

For real solution, we want \(C_{1}>25x^{2}\).

Part b

\begin{align*} \frac{1}{y^{2}}\frac{dy}{dx} & =k\\ \frac{1}{y^{2}}dy & =kdx\\ \frac{-1}{y} & =kx+C\\ y & =\frac{-1}{kx+C} \end{align*}

Part c

\[ \frac{dy}{dx}=1+\frac{y}{x}\qquad x\neq 0 \]

Let \(u=\frac{y}{x}\) or \(y=ux\). Hence \(\frac{dy}{dx}=u^{\prime }x+u\) and the above ODE becomes

\begin{align*} u^{\prime }x+u & =1+u\\ u^{\prime } & =\frac{1}{x}\\ du & =\frac{1}{x}dx\\ u & =\ln \left \vert x\right \vert +C \end{align*}

Hence \[ y=x\left ( \ln \left \vert x\right \vert +C\right ) \]

2.4.6 Problem 5

(a) Solve the IVP \(y^{\prime }\left ( x\right ) =1+4y^{2}\) with \(y\left ( 0\right ) =0\). (b) \(y^{\prime }=-\frac{x}{y}\) with \(y\left ( 1\right ) =\sqrt{3}\) (c) \(e^{x}y^{\prime }=2\left ( x+1\right ) y^{2}\) with \(y\left ( 0\right ) =\frac{1}{6}\)

Part a

\begin{align*} y^{\prime }\left ( x\right ) & =1+4y^{2}\\ \frac{dy}{1+4y^{2}} & =dx\\ \frac{1}{2}\arctan \left ( 2y\right ) & =x+C\\ \arctan \left ( 2y\right ) & =2x+C_{1}\\ y & =\frac{\tan \left ( 2x+C_{1}\right ) }{2} \end{align*}

Applying IC gives

\[ 0=\frac{1}{2}\tan \left ( C_{1}\right ) \]

Hence \(C_{1}=0\). Therefore the solution is  \[ y=\frac{1}{2}\tan \left ( 2x\right ) \]

Part b

\begin{align*} y^{\prime } & =-\frac{x}{y}\\ ydy & =-xdx\\ \frac{1}{2}y^{2} & =-\frac{1}{2}x^{2}+C\\ y^{2} & =-x^{2}+C_{1} \end{align*}

Applying IC gives

\begin{align*} 3 & =-1+C_{1}\\ C_{1} & =4 \end{align*}

Hence solution is \begin{align*} y^{2} & =-x^{2}+4\\ y & =\pm \sqrt{4-x^{2}} \end{align*}

For real solution \(4-x^{2}>0\).

Part c

\begin{align*} e^{x}y^{\prime } & =2\left ( x+1\right ) y^{2}\\ \frac{y^{\prime }}{y^{2}} & =2\left ( x+1\right ) e^{-x}\\ y^{-2}dy & =2\left ( x+1\right ) e^{-x}\\ -\frac{1}{y} & =\int 2\left ( x+1\right ) e^{-x}dx\\ & =-2\left ( x+2\right ) e^{-x}+C \end{align*}

Hence \begin{align*} y & =\frac{1}{2\left ( x+2\right ) e^{-x}+C_{1}}\\ & =\frac{1}{2xe^{-x}+4e^{-x}+C_{1}} \end{align*}

Applying IC gives

\begin{align*} \frac{1}{6} & =\frac{1}{4+C_{1}}\\ 4+C_{1} & =6\\ C_{1} & =2 \end{align*}

Hence solution is

\[ y=\frac{1}{2xe^{-x}+4e^{-x}+2}\]

2.4.7 Key solution