MHB Solving Integrals with Exponential Functions

[bergausstein]

Solve the ff:

$\displaystyle\frac{dy}{dx}-y=xy^5$

$\displaystyle\frac{dy}{dx}-\frac{y}{x}=-\frac{5}{2}x^2y^3$

can you help start solving these problems? thanks!

 

[MarkFL]

These are Bernoulli equations. Such an equation may be written in the form:

$$\frac{dy}{dx}+P(x)y=Q(x)y^n$$

Use of the substitution:

$$v=y^{1-n}$$

will transform the ODE into a linear equation.

 

[bergausstein]

the n=5

$\displaystyle v=y^{1-5}=\frac{1}{y^4}$

$\displaystyle y=(\frac{1}{v})^{\frac{1}{4}}$

$\displaystyle y'=-\frac{1}{4}v^{-\frac{5}{4}}\frac{dv}{dx}$

substitute,

$\displaystyle y=(\frac{1}{v})^{\frac{1}{4}}$ and $\displaystyle y'=-\frac{1}{4}v^{-\frac{5}{4}}\frac{dv}{dx}$ to the original D.E

$\displaystyle -\frac{1}{4}v^{-\frac{5}{4}}\frac{dv}{dx}-\frac{1}{v^{\frac{1}{4}}}=x\frac{1}{v^{\frac{5}{4}}}$

multiplying both sides by $v^{\frac{5}{$}}$ and -4.

$\displaystyle \frac{dv}{dx}+4v=-4x$

the integrating factor is $e^{4x}$

$\displaystyle e^{4x}\frac{dv}{dx}+4ve^{4x}=-4xe^{4x}$

$\displaystyle \int D_x(e^{4x}v)=\int -4xe^{4x}$ the rhs using by parts

$\displaystyle e^{4x}v= -e^{4x}+\frac{1}{4}e^{4x}+c$

dividing both sides by $ e^{4x}$

$\displaystyle v= -1+\frac{1}{4}+\frac{c}{e^{4x}}$

now since $v=\frac{1}{y^4}$

$\displaystyle\frac{1}{y^4}=\frac{c}{e^{4x}}-\frac{3}{4}$ --->> this would be my implicit general solution

is my answer correct?

 

[MarkFL]

We are given:

$$\frac{dy}{dx}-y=xy^5$$

Use the substitution:

$$v=y^{-4}\implies\frac{dv}{dx}= -4y^{-5}\frac{dy}{dx}$$

Divide the original ODE by$y^5$ observing we are eliminating the trivial solution $y\equiv0$. We obtain:

$$y^{-5}\frac{dy}{dx}-y^{-4}=x$$

Applying the substitution we obtain:

$$\frac{dv}{dx}+4v=-4x$$

Multiply through by $e^{4x}$:

$$e^{4x}\frac{dv}{dx}+4e^{4x}v=-4xe^{4x}$$

This becomes:

$$\frac{d}{dx}\left(e^{4x}v \right)=-4xe^{4x}$$

Integrating (using IBP on the right (which you did incorrectly)), we obtain:

$$e^{4x}v=\frac{1}{4}e^{4x}\left(1-4x \right)+C$$

$$v=\frac{1}{4}\left(1-4x \right)+Ce^{-4x}$$

$$y^{-4}=\frac{1}{4}\left(1-4x \right)+Ce^{-4x}$$

Thus the solution may be given implicitly by:

$$y^4=\frac{4}{1-4x+Ce^{-4x}}$$

Note: the trivial solution we eliminated is not obtainable by the implicit solution gave above.

 

[bergausstein]

can you tell me where's my mistake in my solution above?

 

[Prove It]

can you tell me where's my mistake in my solution above?

Edit: There is a small mistake in the integration. Sorry.

\displaystyle \begin{align*} \int{-4x\,e^{4x}\,dx} = -4x\,e^{4x} + \frac{1}{4}e^{4x} + C \end{align*}

 

[bergausstein]

$\displaystyle e^{4x}v= -xe^{4x}+\frac{1}{4}e^{4x}+c$

$\displaystyle v= -x+\frac{1}{4}+\frac{c}{e^{4x}}$

my final answer should be$\displaystyle\frac{1}{y^4}=\frac{c}{e^{4x}}-x+\frac{1}{4}$

am I right?

 

[Prove It]

$\displaystyle e^{4x}v= -xe^{4x}+\frac{1}{4}e^{4x}+c$

$\displaystyle v= -x+\frac{1}{4}+\frac{c}{e^{4x}}$

my final answer should be$\displaystyle\frac{1}{y^4}=\frac{c}{e^{4x}}-x+\frac{1}{4}$

am I right?

Please re-read my last post. You keep missing the factor of -4 on the first term.

 

[bergausstein]

Please re-read my last post. You keep missing the factor of -4 on the first term.

$\displaystyle\int -4xe^{4x}dx=-4\int xe^{4x}$

let $u=x$ and $du=dx$

let $dv=e^{4x}dx$ now $v=\frac{1}{4}e^{4x}$

$uv-\int vdu$

$\frac{1}{4}xe^{4x}-\frac{1}{4}\int e^{4x}dx$

$-4\left(\frac{1}{4}xe^{4x}-\frac{1}{4}\frac{1}{4}e^{4x}+c\right)= -xe^{4x}+\frac{1}{4}e^{4x}+c$

isn't correct?

 

[Prove It]

$\displaystyle\int -4xe^{4x}dx=-4\int xe^{4x}$

let $u=x$ and $du=dx$

let $dv=e^{4x}dx$ now $v=\frac{1}{4}e^{4x}$

$uv-\int vdu$

$\frac{1}{4}xe^{4x}-\frac{1}{4}\int e^{4x}dx$

$-4\left(\frac{1}{4}xe^{4x}-\frac{1}{4}\frac{1}{4}e^{4x}+c\right)= -xe^{4x}+\frac{1}{4}e^{4x}+c$

isn't correct?

Never mind, what you have is correct. I forgot the 1/4 factor in the v term.

I think this many mistakes means it's time for bed >_<