# separable equations (DFQ)

• Jan 19th 2009, 09:31 AM
cheertcc101
separable equations (DFQ)
Alright so I have this equation and I need to solve it using the separable equation method ...

sin(x)dy/dx + ycos(x) = xsin(x)

I tried to separate the equation and got this ...

(1/y)dy = cos(x)/(sin(x)-xsin(x))dx

but I cannot integrate the right side of the equation. I am not sure if I separated them incorrectly or if I am missing something on integrating it.

Any help is appreciated!
• Jan 19th 2009, 09:40 AM
Mush
Quote:

Originally Posted by cheertcc101
Alright so I have this equation and I need to solve it using the separable equation method ...

sin(x)dy/dx + ycos(x) = xsin(x)

I tried to separate the equation and got this ...

(1/y)dy = cos(x)/(sin(x)-xsin(x))dx

but I cannot integrate the right side of the equation. I am not sure if I separated them incorrectly or if I am missing something on integrating it.

Any help is appreciated!

If you divide through the original equation by sin(x) you get:

$\displaystyle \frac{dy}{dx} + y\cot(x) = x$
• Jan 19th 2009, 09:45 AM
cheertcc101
From there how do I get the y and the dy on the same side of the equation so I can integrate??
• Jan 19th 2009, 09:46 AM
ThePerfectHacker
Quote:

Originally Posted by cheertcc101
Alright so I have this equation and I need to solve it using the separable equation method ...

sin(x)dy/dx + ycos(x) = xsin(x)

I tried to separate the equation and got this ...

(1/y)dy = cos(x)/(sin(x)-xsin(x))dx

but I cannot integrate the right side of the equation. I am not sure if I separated them incorrectly or if I am missing something on integrating it.

Any help is appreciated!

Hint: $\displaystyle (y\sin x )' = x\sin x$
• Jan 19th 2009, 10:01 AM
cheertcc101
Alright so the derivative of ysinx = xsinx .. but Im not sure how this helps me at all and also I am not sure that the derivative of ysinx = xsinx so maybe I dont understand what you mean?
• Jan 19th 2009, 10:29 AM
cheertcc101
My bad it was a linear equation and not separable SORRY but THANKS
• Jan 19th 2009, 08:48 PM
Mush
Quote:

Originally Posted by cheertcc101
My bad it was a linear equation and not separable SORRY but THANKS

Your equation is in the form:

$\displaystyle \frac{dy}{dx} + p(x)y = r(x)$

The solution to such an equation is given by:

$\displaystyle \phi(x)y(x) = \int r(x)\phi(x) dx$

Where $\displaystyle \phi(x)$ is the integrating factor. In this case $\displaystyle \phi(x) = e^{\int p(x)dx} = e^{\int \text{cot}(x)dx} =e^{\ln|\sin(x)|} = \sin(x)$

Hence:

$\displaystyle \sin(x)y(x) = \int x\sin(x) dx$

$\displaystyle \sin(x)y(x) = [-x\cos(x)] - \int -cos(x) dx$

$\displaystyle \sin(x)y(x) = -x\cos(x) + \sin(x)+C$

$\displaystyle y(x) = -x\text{cot}(x) + 1+\frac{C}{\sin(x)}$