# Thread: How do i deal with this separable differential equation?

1. ## How do i deal with this separable differential equation?

Hello,

I need to solve this separable differential equation

$4y^3\frac{dy}{dx}=x^{-2}~~~~~,y(1) = 1$

We have $4y^3~dy = \frac{1}{x^2}dx$

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

But what is $\bigg(-\frac{1}{x}\bigg)^{1/4}$?

2. Originally Posted by Jones
Hello,

I need to solve this separable differential equation

$4y^3\frac{dy}{dx}=x^{-2}~~~~~,y(1) = 1$

We have $4y^3~dy = \frac{1}{x^2}dx$

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

But what is $\bigg(-\frac{1}{x}\bigg)^{1/4}$?
The correct answer is $y^4 = - \frac{1}{x} + C$. Then you substitute y(1) = 1 to get the value of C. Then you make y the subject (if that's what the question expects).

3. This is confused,

Don't you want the answer in terms of y?

4. You want y to be the subject usually. ie: y = f(x).

Find C using the way Mr F said: $1 = -\frac{1}{1}+C$

Once the value of C is known you can take the 4th root (obviously substitute in the value of C calculated in the previous step) : $y = \sqrt[4]{-\frac{1}{x}+C}$

As you said you cannot take the 4th root of a negative number so you will have domain restrictions in your answer since $-\frac{1}{x} +C \geq 0$