Solving a Second Order Linear Differential Equation

[Cremolero]

thank you very much!!!!

[georgia1984able]

very nice step-by-step description of the whole procedure.

[pianomanb5]

Since 1 is a double root eigenvalue we only get the basis solution y=e^t. You can show that e^t and te^t are linearly independent since their Wronskian determinant is different from zero. The solution space of this second order D.E. has dimention 2 by the Solution Space Theorem. Thus e^t and te^t form a basis for the vector space of solutions of this equation.

[tir0nel31]

how did you get your homogeneous basis (step 2) to be (e^t, te^t)?? for example if my f(x) the right side of the original equation was (e^x/x) would my homogeneous basis be (e^x, xe^x)??

[Rado3487]

it would be better if you wrote it, while you spoke it would make it easier to follow

[flannelinsummer]

where does that auxiliary condition come from?

[pianomanb5]

we want to solve for v1 and v2, but we only have one condition which is equation 4. If we take the aux condition to be true we now make differentiation easer plus we get 2 equations to solve for 2 unknowns for the linear algebra part. I see you're a musician also. Your guitar riffs sounded good.