1. ## Simple Differential Equation

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

$y(1)=-1$

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}
$

$
y=\frac{4x}{15}-\frac{4x^4}{3}
$

which is totally wrong. What am I missing?

Thank you!

2. 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
$

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

-Dan

3. Thanks. But it still appears to be wrong

4. 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?

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

6. Originally Posted by Vamz
No, I don't have the correct answer. When I derive it, it does not seem to be correct.

7. 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}
$

$
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.