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Calculus

Solve the following problem by separation of variables.
T^2y'-t=ty+y+1
Y(1)=0
Y is a function of t.

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Henry

  1. rearranging things a bit, we have

    y' = y(t+1) + t+1
    y' = (y+1)(t+1)
    dy/(y+1) = (t+1)/t^2 dt
    ln(y+1) = ln(t) - 1/t + c
    y+1 = t e^(-c/t) + c2
    or, altering the c values,
    y = ct e^(-1/t) + c2
    y(1) = 0, so
    0 = c/e + c2, so c2 = -c/e

    y = c(t e^(-1/t) - 1)

    We can ignore the c, since it doesn't really affect anything. Checking, we have

    y' = (t+1)/t e^-(1/t)
    t^2 y' = t(t+1) e^-(1/t)
    Now, we know that e^-(1/t) = (y + 1)/t, so
    t^2 y' = t(t+1)(y + 1)/t
    t^2 y' = (t+1)(y + 1)
    t^2 y' = ty+y+t+1
    t^2 y' - t = ty+y + 1

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    Steve

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