# Math Help - solving for y

1. ## solving for y

dy/dx = 5cos5x/3+2y. The solution I get is (1/2)y^2 + 3y = sin5x + c. However how do I solve for y? Initial condition: y(0) = 1

I have y = (sin5x)/3 -(1/6)(y^2) + 1 Is this legal?

2. Originally Posted by lord12
dy/dx = 5cos5x/3+2y. The solution I get is (1/2)y^2 + 3y = sin5x + c. However how do I solve for y? Initial condition: y(0) = 1

I have y = (sin5x)/3 -(1/6)(y^2) + 1 Is this legal?
as i am feeling lazy today, i'll assume your solution is correct so far. as far as going from $\frac 12y^2 + 3y = \sin 5x + c$ to $y = \cdots$ use the quadratic formula ( $a = \frac 12$, $b = 3$ and $c = \sin 5x + C$)

3. Originally Posted by lord12
dy/dx = 5cos5x/3+2y. The solution I get is (1/2)y^2 + 3y = sin5x + c. However how do I solve for y? Initial condition: y(0) = 1

I have y = (sin5x)/3 -(1/6)(y^2) + 1 Is this legal?
$(3 + 2y) \, dy = 5\cos(5x) \, dx$

$3y + y^2 = \sin(5x) + C$

$y(0) = 1$ ... $C = 4$

$3y + y^2 = \sin(5x) + 4$

$y^2 + 3y + \frac{9}{4} = \sin(5x) + 4 + \frac{9}{4}$

$\left(y + \frac{3}{2}\right)^2 = \sin(5x) + \frac{25}{4}$

$y + \frac{3}{2} = \pm \sqrt{\sin(5x) + \frac{25}{4}}$

since $y(0) = 1$

$y = \sqrt{\sin(5x) + \frac{25}{4}} - \frac{3}{2}$

4. nv