Solve the differential equation

[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.

 

[gulsen]

I'm in a rush, so answers are quite incomplete and tend to be misleading, here they go
1) it seems suitable for y=vx, dy=xdv/dx + v substitution

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

$$y(1-x^2)dx = x(x^2+1)dy$$
$$(x^2-1)dx/x(x^2+1) = dy/y$$
simply integrate both sides.

3) how about giving initial conditions? ;)

 

[prochatz]

Answer for 3)

Hi! I think I have the answer for 3).
Here it goes:

y=c1sin(2x+2) (E.1)

y=c1sin(2x+2) ---> y^2=[c1sin(2x+2)]^2 (E.2)

y=c1sin(2x+2) ---> y'/2=c1cos(2x+2) ---> (y'/2)^2=[c1cos(2x+2)]^2 (E.3)

By adding (E.2) and (E.3) you have:

y^2 + (y'/2)^2=(c1)^2 because (cosk)^2 + (sink)^2 = 1

From (E.1) c1=y/sin(2x+2)

We finally write:

y^2 + (y'/2)^2=[y/sin(2x+2)]^2

 

[hbomb]

for equation 3 shouldn't it be y'/2=4c1cos(2x+c2) sin(2x+c2) for the middle step, because you have to use the chain rule.

 

[HallsofIvy]

First, since this was clearly labled "homework help", I've moved it to the homework section!

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
$$x=5xy+\frac{x^2}{2}+x$$
This makes no sense at all. You appear to have integrated the right hand side with respect to x, treating y as a consant, even though is clearly a function of x, but have not done anything to left hand side. Didn't you learn in algebra "whatever you do to one side of an equation, you must do to the other"?

$$0=5xy+\frac{x^2}{2}$$
$$\frac{-x^2}{2}=5xy$$
$$y=\frac{-x}{10}$$

If this were true, then dy/dx= -1/10. Putting that and y= -x/10 into the original equation gives -x/10= -5x/10+ x+ 1. No, that is not true for all x.

Rewrite the original equation as xdy/dx- 5y= 1- x and then divide by x:
dy/dx- (5/x)y= 1/x- x. That's a linear equation. Do you know how to find an integrating factor for a linear differential equation?

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

Here is what I did:

$$x^2ydx-ydx+x^3dy+xdy=0$$
$$\frac{xdy-ydx}{x^2}+ydx+xdy=0$$

I had to look at this closely: Factoring x² out of x²ydx and x³dy, you get x²(ydx+ xdy) That leaves xdy- ydx so you divide the entire equation by x² to get that.

$$d(\frac{y}{x})+d(xy)=0$$
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.[\quote]
??I don't see why you are stuck. Now, just integrate that equation:
$$\frac{y}{x}+ xy= C$$
Or are you saying you don't see HOW to get that equation? You don't need to multiply by anything:
$$d\frac{y}{x}= \frac{xdy- ydx}{x^2}$$
and
$$d(xy)= ydx+ xdy$$

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.

Since you have two constants, the differential equation must be second order. Yes, y'= 2c₁cos(2x+ c₂) so y"= -4c₁sin(2x+c₂= -4y. That was easy!

 

[hbomb]

Thanks for the help. For question 3 I'm suppose to have a new equation with no c1 or c2 in it. In other words I'm suppose to solve for c1, then take the derivative of the original equation, substitute c1 in for that, then take the derivative of the new equation, then solve for c2. Here is an easy problem to demonstrate what I mean.

$$y=c_1e^{2x}+c_2e^{-x}$$

$$y'=2c_1e^{2x}-c_2e^{-x}$$

$$y"=4c_1e^{2x}+c_2e^{-x}$$

$$y'+y=3c_1e^{2x}$$

$$y"+y'=6c_1e^{2x}$$

$$c_1=\frac{y'+y}{3e^{2x}}$$

$$y"+y'=2y'+2y$$

$$y"-y'-2y=0$$

See...no c1's or c2's