# Thread: Differential equation xy'+y= sinx

1. ## Differential equation xy'+y= sinx

xy'+y= sinx IVP is y(pi)=0

My attempt:

y' +y/x = sinx/x

Finding the I.F

e^$1/x * dx = e^lnx I.F is x Muliply all the equation by x xy' + y = sinx Restate the equation (x*y)' = sinx xy = -cosx + c y = -cosx / x + c I think i have done right so far, but I dont know how to to evaluate it with the values look: 0 = - cos pi / pi + c ??? what How do i evaluate that? 2. -cos(pi)/pi = 1/pi So your equation becomes 0 = 1/pi + c c = -1/pi y = -cos(x)/x - 1/pi (edit) The above corresponds to a previous error in calculation, but I have left it so the following posts don't appear unmotivated 3. It's actually$\displaystyle \displaystyle y = -\frac{\cos{x}}{x} + \frac{c}{x}$... 4. what happened to 1/pi ? how did it become c/ x? didnt get that 5. Originally Posted by Riazy what happened to 1/pi ? how did it become c/ x? didnt get that Prove it just went back to the point right after you integrated$\displaystyle xy=-\cos(x)+C$Now if you isolate y you get$\displaystyle \displaystyle y=-\frac{\cos(x)}{x}+\frac{C}{x}$Now we can use the initial conditions to solve for C$\displaystyle \displaystyle 0=-\frac{\cos(\pi)}{\pi}+\frac{C}{\pi} \implies C=-1\$

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# xy' y=cos x

Click on a term to search for related topics.

#### Search Tags

differential, equation, sinx 