Understanding the General Solution of ODE dy/dx - y = x + 2x^2

[doroulla]

i want to know the general solution of dy/dx - y = x + 2x^2

i don't know how to do it. looked at every book. i can only do it when seperating the variables but here we have "-y"

 

[rock.freak667]

i want to know the general solution of dy/dx - y = x + 2x^2

i don't know how to do it. looked at every book. i can only do it when seperating the variables but here we have "-y"

This is in the form

dy/dx + P(x)y = Q(x)

where P(x) = -1, so just the integrating factor method with P(x) = -1.

 

[doroulla]

i did that but i get integrating factor = e^-x

then i multiplied the original equation with that so i get:

(e^-x)dy/dx - (e^-x)y = (e^-x)(x+2x^2)

but again i can't integrate this

 

[doroulla]

what are the correct steps to do? this is wrong way right? but i followed the books

 

[doroulla]

when i was at university i was really good at this. now i can't seem to figure it out. i have been trying for two days so i decided to join a forum. i never thought someone will reply this fast. please do help me out please

 

[HallsofIvy]

An integrating factor for this equation is a function, u(x) such that
u(x)dy/dx- u(x)y= d(u(x)y)/dx

By the product rule, d(uy)/dx= (du/dx)y+ u(dy/dx) so we want
(du/dx)y+ u(dy/dx)= udu/dx- uy
The "u(dy/dx)" terms cancel leaving y(du/dx)= -uy or du/dx= -u so that u(x)= e^{-x} is, in fact, an integrating factor. Multiplying the equation by that,
e^{-x}(dy/dx)- e^{-x}y= d(e^{-x}y)/dx e^{-x}(x+ 2x^2)[/itex]<br /> The left side integrates to e^{-x}y of course. To integrate the right side, use &quot;integration by parts&quot; taking u= x+ 2x^2 and dv= e^{-x}dx. You will need to use integration by parts twice to complete the integral.<br /> <br /> Another way to do this is to use the fact that this is an &quot;inhomogeneous linear differentiale equation with constant coefficients&quot;. One of the basic properties of &quot;linear&quot; problems in general is that you can solve parts of them and then put those partial solutions together to solve the entire problem.<br /> <br /> Here, you can solve dy/dx- y= 0 easily. You can then try something of the form y= Ax^2+ Bx+ C. Put that into dy/dx- y= x+ 2x^2 and see if you can find values A, B, and C so that function will satify the equation. Add the two solutions to the two different parts to find the general solution to the equation.

 

[doroulla]

thank you very much!
I found general solution is:

y=-2x^2 -5x -5 + C

is this correct? Or am i completely lost? Thank you for your help

 

[HallsofIvy]

You can always check:
If y= -2x^2- 5x- 5+ C then y'= -4x- 5 and then y'- y= -4x- 5+ 2x^2+ 5x+ 5- C= 2x^2+ x- C. No that's not right because of the "C". It would be correct if you took C= 0 but then would not be the "general solution". It's hard to tell if you are "completely lost" because you didn't show how you got that answer. What became of the integrating factor, e^{-t}?

I said before that, because this equation is linear, you could solve y'- y= 0 separately. What do you get as the general solution to that?

 

[doroulla]

i made a mistake. The C should be Ce^x.
Now do you think it makes sense?
I didnt understand the other way so i did it the first way toy explained