# How do i deal with this separable differential equation?

• Jun 19th 2010, 01:45 PM
Jones
How do i deal with this separable differential equation?
Hello,

I need to solve this separable differential equation

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

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

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

But what is $\displaystyle \bigg(-\frac{1}{x}\bigg)^{1/4}$?
• Jun 19th 2010, 01:49 PM
mr fantastic
Quote:

Originally Posted by Jones
Hello,

I need to solve this separable differential equation

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

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

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

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

The correct answer is $\displaystyle y^4 = - \frac{1}{x} + C$. 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).
• Jun 20th 2010, 12:52 AM
Jones
This is confused,

Don't you want the answer in terms of y?
• Jun 20th 2010, 02:48 AM
e^(i*pi)
You want y to be the subject usually. ie: y = f(x).

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

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) : $\displaystyle y = \sqrt[4]{-\frac{1}{x}+C}$

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