1. ## Differential Equation (iii)

I'm doing something wrong, could someone point me to the right direction...

Find the general solution of the differential equation.

$x\frac{dy}{dx}-3y=x^{4}$
ANSWER: $y=x^{3}(x+c)$

---

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

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

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

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

$\int{\frac{3y}{dy}}=\int{\frac{x}{dx}}-\int{x^{4}}$

$\frac{3}{2}y^{2}=\frac{1}{2}x^{2}-\frac{1}{5}x^{5}+c$

...

2. Hello,

Originally Posted by r_maths
I'm doing something wrong, could someone point me to the right direction...

Find the general solution of the differential equation.

$x\frac{dy}{dx}-3y=x^{4}$
ANSWER: $y=x^{3}(x+c)$

---

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

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

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

$\frac{3y}{dy}=\frac{x}{dx}-x^{4}$ SiMoon says : why didn't you divide $x^4$ by dy ?

$\int{\frac{3y}{dy}}=\int{\frac{x}{dx}}-\int{x^{4}}$

SiMoon says : these two last integrals make no sense, you don't know how to integrate when dx or dy is in the denominator

$\frac{3}{2}y^{2}=\frac{1}{2}x^{2}-\frac{1}{5}x^{5}+c$

...

3. How do you divide $x^4$ by dy?

4. Originally Posted by r_maths
I'm doing something wrong, could someone point me to the right direction...

Find the general solution of the differential equation.

$x\frac{dy}{dx}-3y=x^{4}$
ANSWER: $y=x^{3}(x+c)$

---

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

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

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

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

$\int{\frac{3y}{dy}}=\int{\frac{x}{dx}}-\int{x^{4}}$

$\frac{3}{2}y^{2}=\frac{1}{2}x^{2}-\frac{1}{5}x^{5}+c$

...
$x\frac{dy}{dx}-3y=x^{4} \Rightarrow \frac{dy}{dx} - \frac{3}{x} y = x^{3}$.

The integrating factor technique is required.

Note: The integrating factor is $e^{\int -\frac{3}{x} \, dx} = e^{-3 \ln |x|} = e^{\ln |x|^{-3}} = \frac{1}{x^3}$.

5. Originally Posted by r_maths
How do you divide $x^4$ by dy?
You divided it in one term only, which is not logical