# Thread: need someone to solve this question

1. ## need someone to solve this question

2yy' +3 = y^2 + 3x with initial value y(0) = 8

2. ## Re: need someone to solve this question

Originally Posted by mmiraj
2yy' +3 = y^2 + 3x with initial value y(0) = 8
One way to start this is to note that $\displaystyle \left ( y^2 \right ) ' = 2y~y'$, so your equation becomes:
$\displaystyle \left ( y^2 \right )' + 3 = y^2 + 3x$

Now, $\displaystyle y^2$ is just a function, let's call it $\displaystyle z = y^2$, turning your equation into
$\displaystyle z' + 3 = z + 3x$

Can you finish from here?

-Dan

3. ## Re: need someone to solve this question

How is 2yy' = (y^2) ???

4. ## Re: need someone to solve this question

Originally Posted by mmiraj
How is 2yy' = (y^2) ???
It’s not.

topsquark posted $2yy’ = (y^2)’$ ... note the prime symbol after the parentheses, i.e. the derivative of $y^2$ is $2y \cdot y’$

5. ## Re: need someone to solve this question

Originally Posted by topsquark
One way to start this is to note that $\displaystyle \left ( y^2 \right ) ' = 2y~y'$, so your equation becomes:
$\displaystyle \left ( y^2 \right )' + 3 = y^2 + 3x$

Now, $\displaystyle y^2$ is just a function, let's call it $\displaystyle z = y^2$, turning your equation into
$\displaystyle z' + 3 = z + 3x$

Can you finish from here?

-Dan
Let $u = y^2+3x$. Now you have $u' = u$.