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sorry the title should read 2nd order homogenous eq'n, not nonhomogenous

Find the general solution of the equation:

(1+t^2)d^2y/dt^2 - 2t dy/dt + 2y = 0, given that y1(t) = t is one solution.

My attempt:

divided equation by 1+t^2

d^2y/dt^2 - 2t/1+t^2 dy/dt + 2/1+t^2 y = 0

using u(t) eq'n given:

u(t) = exp(-integ(- 2t/t+t^2 dt)) / y1^2(t)

let x = 1+t^2

dx/dt = 2t

dt = dx/2t

u(t) = exp(integ(2t/x dx/2t)) / t^2

u(t) = exp(integ(1/x dx)) / t^2

u(t) = exp(ln x) / t^2

u(t) = e^ln(1+t^2) / t^2

u(t) = e^ln(1+t^2) / t^2

u(t) = (1+t^2) / t^2

u(t) = (1+t^2)/(t^2)

using y2(t) eq'n given:

y2(t) = t integ (u(t)dt)

y2(t) = t integ ((1+t^2)/(t^2)dt)

y2(t) = t [ integ(1/t^2 dt) + integ (1 dt)]

y2(t) = t[-1/t + t]

y2(t) = -1 + t^2

y(t) = c1y1 + c2y2

y(t) = c1t + c2[-1 + t^2]

y(t) = c1t - c2[t^2 - 1]

can someone please confirm whether i am doing this correct?

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# Differential equations - 2nd order nonhomogenous eq'n

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