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2.5 HW 5

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2.5.1 problems description

2.5.2 problem 1

part a

The ODE to solve is

$$\begin{align*}{\frac{\rm d}{{\rm d}x}}y \left ( x \right ) +4\,y \left ( x \right ) =20 \end{align*}$$

with initial conditions $y \left ( 0 \right ) =2$.

Trying separable ODE.

In canonical form, the ODE is written as

$$\begin{align*} y' &= F(x,y)\\ &= -4\,y+20 \end{align*}$$

The ODE $\frac{ \mathop{\mathrm{d}y}}{\mathop{\mathrm{d}x}} = -4\,y+20$, is separable. It can be written as

$$\frac{ \mathop{\mathrm{d}y}}{\mathop{\mathrm{d}x}} = f( x) g(y)$$ Where $f(x)=1$ and $g(y)=-4\,y+20$. Therefore

$$\frac{ \mathop{\mathrm{d}y}}{\mathop{\mathrm{d}x}} = -4\,y+20$$ Hence $$\begin{align*} \left ( -4\,y+20 \right ) ^{-1}\mathop{\mathrm{d}y}&= \mathop{\mathrm{d}x}\\ \int \left ( -4\,y+20 \right ) ^{-1}\mathop{\mathrm{d}y}&= \int \mathop{\mathrm{d}x}\\ -1/2\,\ln \left ( 2 \right ) -1/4\,\ln \left ( \left | y-5 \right | \right ) &=x+C_{{1}}\\ \end{align*}$$

Solving for $y$ gives

$$y = -1/4\,{{\rm e}^{-4\,x-4\,C_{{1}}}}+5$$ The solution above can be written as $$\begin{equation} \setlength{\fboxsep }{2.5\fboxsep }\boxed{y = -1/4\,C_{{1}}{{\rm e}^{-4\,x}}+5} \end{equation}$$

Initial conditions are now used to solve for $C_{{1}}$. Substituting $x=0$ and $y=2$ in the above solution gives an equation to solve for the constant of integration.

$$\begin{align*} 2&= -1/4\,C_{{1}}{{\rm e}^{0}}+5\\ &=-1/4\,C_{{1}}+5 \end{align*}$$

Hence

$$C_{{1}} = 12\, \left ({{\rm e}^{0}} \right ) ^{-1}$$ Which is simplified to $$C_{{1}} = 12$$ Substituting $C_1$ found above back in the solution gives $$y \left ( x \right ) = -3\,{{\rm e}^{-4\,x}}+5$$

part b

The ODE to solve is

$$\begin{align*}{\frac{\rm d}{{\rm d}x}}y \left ( x \right ) +3\,y \left ( x \right ) =\sin \left ( x \right ) \end{align*}$$

with initial conditions $y \left ( \pi /2 \right ) =3/10$.

Trying Linear ODE.

In canonical form, the ODE is written as

$$\begin{align*} y' &= F(x,y)\\ &= -3\,y+\sin \left ( x \right ) \end{align*}$$

The ODE is linear in $y$ and has the form

$$y' = y f(x) + g(x)$$ Where $f(x) = -3$ and $g(x) = \sin \left ( x \right )$.

Writing the ODE as

$$\begin{align*} y' - \left (-3\,y\right ) &= \sin \left ( x \right ) \\ y' + 3\,y &= \sin \left ( x \right ) \end{align*}$$

Therefore the integrating factor $\mu$ is

$$\mu = e^{\int 3\mathop{\mathrm{d}x}} ={{\rm e}^{3\,x}}$$ The ode becomes $$\begin{align*} \frac{\mathop{\mathrm{d}}}{ \mathop{\mathrm{d}x}} \mu y &= \mu \left (\sin \left ( x \right ) \right ) \\ \frac{\mathop{\mathrm{d}}}{ \mathop{\mathrm{d}x}} \left (y{{\rm e}^{3\,x}}\right ) &= \sin \left ( x \right ){{\rm e}^{3\,x}}\\ \mathrm{d} \left (y{{\rm e}^{3\,x}}\right ) &= \left (\sin \left ( x \right ){{\rm e}^{3\,x}}\right ) \mathrm{d} x \end{align*}$$

Integrating both sides gives

$$\begin{align*} y{{\rm e}^{3\,x}} &= -1/10\,\cos \left ( x \right ){{\rm e}^{3\,x}}+3/10\,\sin \left ( x \right ){{\rm e}^{3\,x}} + C_1 \end{align*}$$

Dividing both sides by the integrating factor $\mu ={{\rm e}^{3\,x}}$ results in

$$y ={\frac{-1/10\,\cos \left ( x \right ){{\rm e}^{3\,x}}+3/10\,\sin \left ( x \right ){{\rm e}^{3\,x}}}{{{\rm e}^{3\,x}}}}+{\frac{C_{{1}}}{{{\rm e}^{3\,x}}}}$$ Simplifying the solution gives $$\setlength{\fboxsep }{2.5\fboxsep }\boxed{y=3/10\,\sin \left ( x \right ) -1/10\,\cos \left ( x \right ) +C_{{1}}{{\rm e}^{-3\,x}}}$$

Initial conditions are now used to solve for $C_{{1}}$. Substituting $x=\pi /2$ and $y=3/10$ in the above solution gives an equation to solve for the constant of integration.

$$\begin{align*} 3/10&= 3/10\,\sin \left ( \pi /2 \right ) -1/10\,\cos \left ( \pi /2 \right ) +C_{{1}}{{\rm e}^{-3/2\,\pi }}\\ &=3/10+C_{{1}}{{\rm e}^{-3/2\,\pi }} \end{align*}$$

Hence

$$C_{{1}} = -1/10\,{\frac{3\,\sin \left ( \pi /2 \right ) -\cos \left ( \pi /2 \right ) -3}{{{\rm e}^{-3/2\,\pi }}}}$$ Which is simplified to $$C_{{1}} = 0$$ Substituting $C_1$ found above back in the solution gives $$y \left ( x \right ) = 3/10\,\sin \left ( x \right ) -1/10\,\cos \left ( x \right )$$

part c

The ODE to solve is

$$\begin{align*}{\frac{\rm d}{{\rm d}x}}y \left ( x \right ) -y \left ( x \right ) \left ( 1+3\,{x}^{-1} \right ) =x+2 \end{align*}$$

with initial conditions $y \left ( 1 \right ) ={\rm e}-1$.

Trying Linear ODE.

In canonical form, the ODE is written as

$$\begin{align*} y' &= F(x,y)\\ &={\frac{{x}^{2}+xy+2\,x+3\,y}{x}} \end{align*}$$

The ODE is linear in $y$ and has the form

$$y' = y f(x) + g(x)$$ Where $f(x) ={\frac{x+3}{x}}$ and $g(x) ={\frac{{x}^{2}+2\,x}{x}}$.

Writing the ODE as

$$\begin{align*} y' - \left ({\frac{ \left ( x+3 \right ) y}{x}}\right ) &={\frac{{x}^{2}+2\,x}{x}}\\ y' -{\frac{ \left ( x+3 \right ) y}{x}} &={\frac{{x}^{2}+2\,x}{x}} \end{align*}$$

Therefore the integrating factor $\mu$ is

$$\mu = e^{\int -{\frac{x+3}{x}}\mathop{\mathrm{d}x}} ={{\rm e}^{-x-3\,\ln \left ( x \right ) }}$$ The ode becomes $$\begin{align*} \frac{\mathop{\mathrm{d}}}{ \mathop{\mathrm{d}x}} \mu y &= \mu \left ({\frac{{x}^{2}+2\,x}{x}}\right ) \\ \frac{\mathop{\mathrm{d}}}{ \mathop{\mathrm{d}x}} \left (y{{\rm e}^{-x-3\,\ln \left ( x \right ) }}\right ) &={\frac{ \left ({x}^{2}+2\,x \right ){{\rm e}^{-x-3\,\ln \left ( x \right ) }}}{x}}\\ \mathrm{d} \left (y{{\rm e}^{-x-3\,\ln \left ( x \right ) }}\right ) &= \left ({\frac{ \left ({x}^{2}+2\,x \right ){{\rm e}^{-x-3\,\ln \left ( x \right ) }}}{x}}\right ) \mathrm{d} x \end{align*}$$

Integrating both sides gives

$$\begin{align*} y{{\rm e}^{-x-3\,\ln \left ( x \right ) }} &= -{{\rm e}^{-x-3\,\ln \left ( x \right ) }}x + C_1 \end{align*}$$

Dividing both sides by the integrating factor $\mu ={{\rm e}^{-x-3\,\ln \left ( x \right ) }}$ results in

$$y = -x+{\frac{C_{{1}}}{{{\rm e}^{-x-3\,\ln \left ( x \right ) }}}}$$ Simplifying the solution gives $$\setlength{\fboxsep }{2.5\fboxsep }\boxed{y=-x+C_{{1}}{x}^{3}{{\rm e}^{x}}}$$

Initial conditions are now used to solve for $C_{{1}}$. Substituting $x=1$ and $y={\rm e}-1$ in the above solution gives an equation to solve for the constant of integration.

$$\begin{align*}{\rm e}-1&=-1+C_{{1}}{\rm e} \end{align*}$$

Hence

$$C_{{1}} = 1$$ Substituting $C_1$ found above back in the solution gives $$y \left ( x \right ) = -x+{x}^{3}{{\rm e}^{x}}$$

2.5.3 problem 2

The ODE to solve is

$$\begin{align*}{\frac{\rm d}{{\rm d}x}}y \left ( x \right ) +1/3\,y \left ( x \right ) =1/3\, \left ( 1-2\,x \right ) \left ( y \left ( x \right ) \right ) ^{4} \end{align*}$$

Trying Bernoulli ODE.

In canonical form, the ODE is written as

$$\begin{align*} y' &= F(x,y)\\ &= -y/3-2/3\,{y}^{4}x+1/3\,{y}^{4} \end{align*}$$

This is a Bernoulli ODE. Comparing the ODE to solve

$$y' = -y/3-2/3\,{y}^{4}x+1/3\,{y}^{4}$$ With Bernoulli ODE standard form $$y' = f_0(x)y+f_1(x)y^n$$

Shows that $f_0(x)=-1/3$ and $f_1(x)=-2/3\,x+1/3$ and $n=4$.

Dividing the ODE by ${y}^{4}$ gives

$$\begin{align*} y'{y}^{-4} &= -1/3{y}^{-3} + -2/3\,x+1/3 \tag{1} \end{align*}$$

Let

$$\begin{align*} v &={y}^{-3} \tag{2} \end{align*}$$

Taking derivative of (2) w.r.t $x$ gives

$$\begin{align*} v' &= -3\,{y}^{-4}y' \\{y}^{-4}&= \frac{v'}{-3\,y'} \tag{3} \end{align*}$$

Substituting (3) into (1) gives

$$\begin{align*} \frac{v'}{(-3)} &= \left (-1/3\right ) v + -2/3\,x+1/3\\ v' &= (-3)\left (-1/3\right ) v + (-3)\left (-2/3\,x+1/3\right ) \\ &= v+2\,x-1 \end{align*}$$

The above now is a linear ODE in $v(x)$ which can be easily solved using an integrating factor.

In canonical form, the ODE is written as

$$\begin{align*} v' &= F(x,v)\\ &= v+2\,x-1 \end{align*}$$

The ODE is linear in $v$ and has the form

$$v' = v f(x) + g(x)$$ Where $f(x) = 1$ and $g(x) = 2\,x-1$.

Writing the ODE as

$$\begin{align*} v' - \left (v\right ) &= 2\,x-1\\ v' -v &= 2\,x-1 \end{align*}$$

Therefore the integrating factor $\mu$ is

$$\mu = e^{\int -1\mathop{\mathrm{d}x}} ={{\rm e}^{-x}}$$ The ode becomes $$\begin{align*} \frac{\mathop{\mathrm{d}}}{ \mathop{\mathrm{d}x}} \mu v &= \mu \left (2\,x-1\right ) \\ \frac{\mathop{\mathrm{d}}}{ \mathop{\mathrm{d}x}} \left (v{{\rm e}^{-x}}\right ) &= \left ( 2\,x-1 \right ){{\rm e}^{-x}}\\ \mathrm{d} \left (v{{\rm e}^{-x}}\right ) &= \left ( \left ( 2\,x-1 \right ){{\rm e}^{-x}}\right ) \mathrm{d} x \end{align*}$$

Integrating both sides gives

$$\begin{align*} v{{\rm e}^{-x}} &= - \left ( 2\,x+1 \right ){{\rm e}^{-x}} + C_1 \end{align*}$$

Dividing both sides by the integrating factor $\mu ={{\rm e}^{-x}}$ results in

$$v = -2\,x-1+{\frac{C_{{1}}}{{{\rm e}^{-x}}}}$$ Simplifying the solution gives $$\setlength{\fboxsep }{2.5\fboxsep }\boxed{v=-2\,x-1+C_{{1}}{{\rm e}^{x}}}$$

Replacing $v$ in the above by ${y}^{-3}$ from equation (2), gives the final solution.

$${y}^{-3}=-2\,x-1+C_{{1}}{{\rm e}^{x}}$$ Solving for $y$ gives $$\begin{align*} y &={\frac{1}{\sqrt [3]{-2\,x-1+C_{{1}}{{\rm e}^{x}}}}}\\ y &=-1/2\,{\frac{1}{\sqrt [3]{-2\,x-1+C_{{1}}{{\rm e}^{x}}}}}+{\frac{i/2\sqrt{3}}{\sqrt [3]{-2\,x-1+C_{{1}}{{\rm e}^{x}}}}}\\ y &=-1/2\,{\frac{1}{\sqrt [3]{-2\,x-1+C_{{1}}{{\rm e}^{x}}}}}-{\frac{i/2\sqrt{3}}{\sqrt [3]{-2\,x-1+C_{{1}}{{\rm e}^{x}}}}}\\ \end{align*}$$

2.5.4 problem 3

The ODE to solve is

$$\begin{align*} m{\frac{\rm d}{{\rm d}x}}v \left ( x \right ) =w-B-kv \left ( x \right ) \end{align*}$$

with initial conditions $v \left ( 0 \right ) =0$.

Trying separable ODE.

In canonical form, the ODE is written as

$$\begin{align*} v' &= F(x,v)\\ &= -{\frac{kv+B-w}{m}} \end{align*}$$

The ODE $\frac{ \mathop{\mathrm{d}v}}{\mathop{\mathrm{d}x}} = -{\frac{kv+B-w}{m}}$, is separable. It can be written as

$$\frac{ \mathop{\mathrm{d}v}}{\mathop{\mathrm{d}x}} = f( x) g(v)$$ Where $f(x)=1$ and $g(v)={\frac{-kv-B+w}{m}}$. Therefore

$$\frac{ \mathop{\mathrm{d}v}}{\mathop{\mathrm{d}x}} ={\frac{-kv-B+w}{m}}$$ Hence $$\begin{align*} \left ({\frac{m}{-kv-B+w}}\right )\mathop{\mathrm{d}v}&= \mathop{\mathrm{d}x}\\ \int \left ({\frac{m}{-kv-B+w}}\right )\mathop{\mathrm{d}v}&= \int \mathop{\mathrm{d}x}\\ -{\frac{m\ln \left ( \left | kv+B-w \right | \right ) }{k}}&=x+C_{{1}}\\ \end{align*}$$

Solving for $v$ gives

$$v ={\frac{1}{k} \left ( -{{\rm e}^{-{\frac{k \left ( x+C_{{1}} \right ) }{m}}}}-B+w \right ) }$$

Initial conditions are now used to solve for $C_{{1}}$. Substituting $x=0$ and $v=0$ in the above solution gives an equation to solve for the constant of integration.

$$\begin{align*} 0&={\frac{1}{k} \left ( -{{\rm e}^{-{\frac{kC_{{1}}}{m}}}}-B+w \right ) } \end{align*}$$

Hence

$$C_{{1}} = -{\frac{m\ln \left ( -B+w \right ) }{k}}$$ Substituting $C_1$ found above back in the solution gives $$v \left ( x \right ) ={\frac{1}{k} \left ( -{{\rm e}^{-{\frac{k}{m} \left ( x-{\frac{m\ln \left ( -B+w \right ) }{k}} \right ) }}}-B+w \right ) }$$

The solution $\frac{1}{k} \left ( -{{\rm e}^{-{\frac{k}{m} \left ( x-{\frac{m\ln \left ( -B+w \right ) }{k}} \right ) }}}-B+w \right )$ can be simplified to

$$\begin{equation} \setlength{\fboxsep }{2.5\fboxsep }\boxed{v \left ( x \right ) ={\frac{1}{k} \left ( -{{\rm e}^{{\frac{m\ln \left ( -B+w \right ) -xk}{m}}}}-B+w \right ) }} \end{equation}$$

2.5.5 problem 4

The ODE to solve is

$$\begin{align*}{\frac{\rm d}{{\rm d}x}}y \left ( x \right ) ={x}^{3} \left ( y \left ( x \right ) -x \right ) ^{2}+{\frac{y \left ( x \right ) }{x}} \end{align*}$$

Trying Riccati ODE.

In canonical form, the ODE is written as

$$\begin{align*} y' &= F(x,y)\\ &={\frac{{x}^{6}-2\,{x}^{5}y+{x}^{4}{y}^{2}+y}{x}} \end{align*}$$

This is a Riccati ODE. Comparing the ODE to solve

$$y' ={x}^{5}-2\,{x}^{4}y+{x}^{3}{y}^{2}+{\frac{y}{x}}$$ With Riccati ODE standard form $$y' = f_0(x)+ f_1(x)y+f_2(x){y}^{2}$$

Shows that $f_0(x)={x}^{5}$, $f_1(x)={\frac{-2\,{x}^{5}+1}{x}}$ and $f_2(x)={x}^{3}$.

Let

$$\begin{align*} y &= \frac{-u'}{f_2 u} \\ &= \frac{-u'}{u{x}^{3}} \tag{1} \end{align*}$$

Using the above substitution in the given ODE results (after some simplification) in a second order ODE to solve for $u(x)$ which is

$$\begin{align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag{2} \end{align*}$$

But

$$\begin{align*} f_2' &=3\,{x}^{2}\\ f_1 f_2 &= \left ( -2\,{x}^{5}+1 \right ){x}^{2}\\ f_2^2 f_0 &={x}^{11} \end{align*}$$

Substituting the above terms back in (2) gives

$$\begin{align*}{x}^{3}{\frac{{\rm d}^{2}}{{\rm d}{x}^{2}}}u \left ( x \right ) - \left ( 3\,{x}^{2}+ \left ( -2\,{x}^{5}+1 \right ){x}^{2} \right ){\frac{\rm d}{{\rm d}x}}u \left ( x \right ) +{x}^{11}u \left ( x \right ) &=0 \end{align*}$$

Solving the above ODE gives

$$u \left ( x \right ) ={{\rm e}^{-1/5\,{x}^{5}}} \left ({x}^{5}C_{{2}}+C_{{1}} \right )$$ The above shows that $$u'(x) = -{x}^{4}{{\rm e}^{-1/5\,{x}^{5}}} \left ({x}^{5}C_{{2}}+C_{{1}}-5\,C_{{2}} \right )$$ Hence, using the above in (1) gives the solution $$y \left ( x \right ) ={\frac{x \left ({x}^{5}C_{{2}}+C_{{1}}-5\,C_{{2}} \right ) }{{x}^{5}C_{{2}}+C_{{1}}}}$$ Dividing both numerator and denominator by $C_2$ gives, after renaming the constant $\frac{C_1}{C_2}=C_0$ the following

$$\boxed{y \left ( x \right ) ={\frac{x \left ({x}^{5}+C_{{0}}-5 \right ) }{{x}^{5}+C_{{0}}}}}$$

2.5.6 Key solution