Differential equation problems....

antiflag403

Hey everyone,
Having some problems with these differential equations. If someone could point me in the right direction I would be really grateful.
1) dy/dx=(5x+12y)/3x with initial conditions y(1)=5.
I know this is a seperable equation but I'm not sure of the correct algebra to seperate them. Do I make it 5x/3x +12y/3x and then work from there??
2) dx/dt+3x=cos(2t) x(0)=-1
I know this one is a first order linear differential equation so it must be in the form dx/dt+p(x)y=q(x) and then you take e^(integral)p(x) and so on. I'm not sure what to do since there is no p(x) (or is it just 3??? integral being 3x). I tried that but I have trouble with the integration by parts that results.
If anyone could give me some advice or point me in the right direction that would be great.
Thanks

HallsofIvy

Then you do not know what a "separable" equation is! How would you propose to "separate" x and y?

No, it's not of the form dx/dt+ p(x)y= q(x)!! It is of the form
dx/dt+ p(t)x= q(t). And then, Yes, p(t)= 3.
[tex]\int e^{3t}cos(2t)dt[/tex] requires integration by parts twice and then think!!

antiflag403

well if the first one is not seperable then I don't know of a way to solve it. Could you give me a hint. Thanks for the help by the way.

pizzasky

Here's some help for the 1st differential equation...

Some algebraic manipulation will give you the expression

[tex] \frac{dy}{dx} - \frac{4}{x} y = \frac{5}{3} [/tex]

Do you know how to solve it now?

All the best!

arildno

The first is, in fact, separable, if you make a change of variables:
[tex]\frac{dy}{dx}=\frac{5x+12y}{3x}=\frac{5}{3}+4\frac{y}{x}=f(\frac{y}{x})[/tex]

Introduce the variable:
[tex]u=\frac{y}{x}\to\frac{du}{dx}=\frac{\frac{dy}{dx}-u}{x}\to\frac{1}{(f(u)-u)}\frac{du}{dx}=\frac{1}{x}[/tex]

In this case, of course, we have: [itex]\frac{dy}{dx}=f(u)=\frac{5}{3}+4u[/itex]