# Math Help - separable equations (DFQ)

1. ## 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!

2. 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:

$\frac{dy}{dx} + y\cot(x) = x$

3. From there how do I get the y and the dy on the same side of the equation so I can integrate??

4. 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: $(y\sin x )' = x\sin x$

5. 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?

6. My bad it was a linear equation and not separable SORRY but THANKS

7. Originally Posted by cheertcc101
My bad it was a linear equation and not separable SORRY but THANKS
Your equation is in the form:

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

The solution to such an equation is given by:

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

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

Hence:

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

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

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

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