First oreder linear ODE NONHOMOGENEOUS

[pat666]

Homework Statement

$$x^3y'+xy=x, y(1)=2$$

Homework Equations

The Attempt at a Solution

I'm having trouble starting this because it doesn't fit any form I'm familiar with because of the $$x^3$$ in front of the y'. Can someone give me some pointers to get started..

 

[fzero]

Unless you have a typo, you can simplify the equation somewhat by dividing through by x. The homogeneous equation is separable. The inhomogenous equation should be solvable by variation of parameters or by just guessing that the particular solution is a constant.

 

[pat666]

hey, thanks for the reply. I'm still having trouble though.
$$x^2y'=1-y$$
$$x^2/dx=(1-y)/dy$$
I'm not sure what to do now because I can't integrate like that?

Thanks

edit:
unless I should invert everything:
$$1/x^2 dx=1/(1-y) dy$$ just an idea not sure if that is the logical next step?

 

[Dick]

hey, thanks for the reply. I'm still having trouble though.
$$x^2y'=1-y$$
$$x^2/dx=(1-y)/dy$$
I'm not sure what to do now because I can't integrate like that?

Thanks

edit:
unless I should invert everything:
$$1/x^2 dx=1/(1-y) dy$$ just an idea not sure if that is the logical next step?

Yes, invert both sides.

 

[pat666]

ok now I have
$$1/x+c=ln(y-1)$$
$$y=e^(1/x)+1+c_1$$

Is that correct?
Thanks

 

[Dick]

Integrating dx/x^2 doesn't give you 1/x. And you need to think a bit harder about what happens when you exponentiate both sides.

 

[pat666]

it gives -1/x but the integral of $$1/(1-y) dy is -ln(y-1)$$ so I got rid of the negative. I don't know what's wrong with the exponentiation though?? unless you meant that c should be e^c but I changed to c1 where c1=e^cthanks

 

[HallsofIvy]

$$e^{(1/x)+ C}= e^C e^{1/x}= c_1e^{1/x}$$
not
"$$e^{1/x}+ c_1$$"

 

[pat666]

ok thanks so:
$$y=c_1 e^(1/x)+1$$ is a/the correct solutions?

thanks

 

[Dick]

ok thanks so:
$$y=c_1 e^(1/x)+1$$ is a/the correct solutions?

thanks

That's it.