Finding two solutions for the d.e y''=1

[laura_a]

Homework Statement

I need to find the solution to the d.e y''=1 with initial conditions y(0)=y'(0)=0

I have the formula to use
Y(x) = y_2(x) \int \frac{y_1}{W(t)} g(t)dt - y_1(x) \int \frac{y_2}{W(t)}g(t)dt

And I've worked out easily that one solution is \frac{x^2}{2}
I just tried to use the formula I've got to find another solution y_2 but I ended up getting the same as I've got for y_1.

Is there an easier way to find a solution. The first part of the question asked to find 2 solutions to the d.e y''=0 and I found them to be y_1=1 and y_2=x, it says I can use this information to solve the d.e y''=1 but I don't see how?

Can anyone lead me in the right direction? :) Thanks

 

[Defennder]

You need one particular solution and two solutions from the homogenous DE. Seems you've got them all. What's wrong?

 

[HallsofIvy]

It's hardly that complicated. If u'= 1 the "slope" is a constant so u is linear and u(x)= x+ C where C is any constant.

If you set u= y', y"= u'= 1 so u= y'= x+ C. Now integate again: y= (1/2)x²+ Cx +D.

Taking different values for C and D will give you an infinite number of solutions but the only one that satisfies both y(0)= 0 and y'(0)= 0 is y= (1/2)x².

If you are concerned about a second order d.e. having two independent solutions, for y"= 0, they are y= 1 and y= x: integrating twice gives y= Cx+ D and then taking C= 1, D= 0 gives y= 1 while taking C= 0, D= 1 gives y= x.

Equivalently, the characteristic equation for the associated homogeneous equation is r²= 0 which has r= 0 as a double root. The independent solutions to the homogeneous equation are y= e^0x= 1 and y= xe^0x= x.