Solve the differential equation xdy=(5y+x+1)dx

[hbomb]

I need someone to look over my work if possible

1) Solve the differential equation
xdy=(5y+x+1)dx

Here is what I did:
x=(5y+x+1)dx/dy
x=5xy+\frac{x^2}{2}+x
0=5xy+\frac{x^2}{2}
\frac{-x^2}{2}=5xy
y=\frac{-x}{10}

2) Solve: y(x^2-1)dx+x(x^2+1)dy=0

Here is what I did:

x^2ydx-ydx+x^3dy+xdy=0

\frac{xdy-ydx}{x^2}+ydx+xdy=0

d(\frac{y}{x})+d(xy)=0
This is where I am stuck at. I need to have something multiply the first differential that will yield something in the form of \frac{y}{x} and also have the same thing multiply the second differential and will yield something in the form of xy.

3) Find a differential equation with the solution y=c_1sin(2x+c_2)

Here is what I did:

y'=2c_1cos(2x+c_2)

and since
c_1=\frac{y}{sin(2x+c_2)}

y'=\frac{2ycos(2x+c_2)}{sin(2x+c_2}

y'=2ycot(2x+c_2)

I'm not sure what to do after this. I know I need to somehow get rid of the c_2 constant, but I don't know how to do this.

 

[Benny]

In the second one just get all of the x and dx terms on one side and all of the y and dy terms on the other. That should only take one step.

<br /> y\left( {x^2 - 1} \right)dx + x\left( {x^2 + 1} \right)dy = 0<br />

<br /> \Rightarrow \left( {\frac{{x^2 - 1}}{{x^2 + 1}}} \right)dx = - \frac{1}{y}dy<br />

Now integrate both sides.