Simple Differential Equation

**Vamz** · March 21st 2011, 05:37 PM

$\frac{dy}{dx} = \frac{4 x + 20 y}{5 x}$

$y(1)=-1$

What did I do wrong here?
$y'=\frac{4x}{5x}+\frac{20y}{5x}$

$y'-\frac{20}{5}x^{-1}y=\frac{4x}{5x}$ < Is this proper first order linear?

let $\eta=e^{\int\frac{20}{5x}}= e^{-4ln(x)}$

Mulitplay eta on both sides, and integrate, I get:

$ yx^{-4}=\frac{4}{5}\int x^{-4} dx $

$ y=\frac{4x}{15}+x^4C $

To solve for C:
$ -1=\frac{4}{15}+C, C= -\frac{4}{3} $

My answer is:
$ y=\frac{4x}{15}-\frac{4x^4}{3} $

which is totally wrong. What am I missing?

Thank you!

**topsquark** · March 21st 2011, 05:44 PM

Originally Posted by Vamz

$\frac{dy}{dx} = \frac{4 x + 20 y}{5 x}$

$y(1)=-1$

What did I do wrong here?
$y'=\frac{4x}{5x}+\frac{20y}{5x}$

$y'-\frac{20}{5}x^{-1}y=\frac{4x}{5x}$ < Is this proper first order linear?

let $\eta=e^{\int\frac{20}{5x}}= e^{-4ln(x)}$

Mulitplay eta on both sides, and integrate, I get:

$ yx^{-4}=\frac{4}{5}\int x^{-4} dx $

$ y=\frac{4x}{15}+x^4C $

I don't know about totally wrong. You missed a minus sign in this last line:
$ y=-\frac{4x}{15}+x^4C $

-Dan

**Vamz** · March 21st 2011, 05:48 PM

Thanks. But it still appears to be wrong

**Prove It** · March 21st 2011, 05:49 PM

Originally Posted by Vamz

Thanks. But it still appears to be wrong

Have you been told what the correct answer is? If not, have you tried differentiating and substituting back into the DE to see if the DE is satisfied?

**Vamz** · March 21st 2011, 06:10 PM

No, I don't have the correct answer. When I derive it, it does not seem to be correct.

**Prove It** · March 21st 2011, 06:14 PM

Originally Posted by Vamz

No, I don't have the correct answer. When I derive it, it does not seem to be correct.

Then like I said, substitute your answer and your derivative into your DE. Find out if the solution satisfies the DE...

**dwsmith** · March 21st 2011, 06:40 PM

Originally Posted by Vamz

$\frac{dy}{dx} = \frac{4 x + 20 y}{5 x}$

$y(1)=-1$

What did I do wrong here?
$y'=\frac{4x}{5x}+\frac{20y}{5x}$

$y'-\frac{20}{5}x^{-1}y=\frac{4x}{5x}$ < Is this proper first order linear?

let $\eta=e^{\int\frac{20}{5x}}= e^{-4ln(x)}$

Mulitplay eta on both sides, and integrate, I get:

$ yx^{-4}=\frac{4}{5}\int x^{-4} dx $

$ y=\frac{4x}{15}+x^4C $

To solve for C:
$ -1=\frac{4}{15}+C, C= -\frac{4}{3} $

My answer is:
$ y=\frac{4x}{15}-\frac{4x^4}{3} $

which is totally wrong. What am I missing?

Thank you!

$\displaystyle 5xy'=4+20y\Rightarrow 5xy'-20y=4\Rightarrow y'-\frac{4y}{x}=\frac{4}{5}$

$\displaystyle\frac{y}{x^4}=\int \frac{4}{5}x^{-4} \ dx\Rightarrow yx^{-4}=\frac{-4x^{-3}}{15}+C$

$\displaystyle y=Cx^4-\frac{4x}{15}$

$\displaystyle f(1)=C-\frac{4}{15}=-1\Rightarrow C=-\frac{11}{15}$

This is what I obtained.

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