For the first order linear differential equation dy / dt + 2t / t2 + 1 y = t, De

Question

For the first order linear differential equation dy / dt + 2t / t2 + 1 y = t, Define a new dependent variable u(t) = y(t)(t2 + 1). Express y(t) and dy/dt in terms of u(t) and du/dt. Substitute your formulas from (c) into the original differential equation to get a new differential equation for u(t). I don't mind telling you that the general solution to your diff eq in (b) is u(t) = t4 / 4 + t2 / 2 + c, where C is an arbitrary constant. Use this to find the general solution y(t) for the original diff eq.. Which are the complementary and particular parts of the solution in (c)?

Explanation / Answer

a)

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

so,

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

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

diferentiate we get

du/dt = dy/dt * (t^2+1) + y(t)*2t

or

dy/dt* (t^2+1) = du/dt - 2ty(t)

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

we get

dy/dt * (t^2+1) = du/dt - 2t * u(t)/(t^2+1)

or

dy/dt = du/dt *1/(t^2+1) - 2t * u(t)/(t^2+1)^2

which is in the expression of dy/dt in terms of u(t) and du/dt.

b)

put dy/dt = du/dt *1/(t^2+1) - 2t * u(t)/(t^2+1)^2 in the gicen diferential eq. we get

du/dt *1/(t^2+1) - 2t * u(t)/(t^2+1)^2 + (2t/(t^2+1)) * u(t)/(t^2+1)

= t

or

du/dt *1/(t^2+1) = t

or

du = (t^3 + t)dt

solution is

u(t) = t^4/4 + t^2/2 + C

we know u(t) = y(t)(t^2+1)

so, we get

y(t)(t^2+1) = t^4/4 + t^2/2 + C

or

y(t) = [t^4/4 + t^2/2 + C ]/[t^2+1]

which is the solution of the original diffeential eq.

d)

complementary part is = t^4/4 + t^2/2

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