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# Find the solution of the given initial value problem?

Find the solution of the given initial value problem for ty'+2y=t^2-t+1, y(1)=1/2, t>0.

### 1 Answer by Expert Tutors

Roman C. | Masters of Education Graduate with Mathematics ExpertiseMasters of Education Graduate with Mathe...
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Again, the integrating factor comes to use.

t dy/dt + 2y = t2 - t + 1

dy/dt + 2y/t = = t - 1 + 1/t

μ dy/dt + 2μy/t = μ(t - 1 + 1/t)

Let dμ/dt = 2μ/t.

μ dy/dt + y dμ/dt = μ(t - 1 + 1/t)

μy = ∫ μ(t - 1 + 1/t) dt

y = (1/μ) ∫ μ(t - 1 + 1/t) dt

Find μ from it's equation.

dμ/dt = 2μ/t

∫ dμ/μ = ∫ 2dt/t

ln |μ| = 2 ln |t| + C = ln t2 + C

μ = At2

Plug μ in.

y = (1/(At2)) ∫ At2(t - 1 + 1/t) dt

y = (1/t2) ∫ (t3 - t2 + t) dt

y = (t4/4 - t3/3 + t2/2 + C) / t2

y = t2/4 - t/3 + 1/2 + C/t2

Before using the initial value, let's check the general solution.

t * (d/dt)(t2/4 - t/3 + 1/2 + C/t2) + 2(t2/4 - t/3 + 1/2 + C/t2)

= t * (t/2 - 1/3 + 0 - 2C/t3) + 2(t2/4 - t/3 + 1/2 + C/t2)

= t2/2 - t/3 - 2C/t2 + t2/2 - 2t/3 + 1 + 2C/t2

= t2 - t + 1

It checks out.

Now since we have that y = 1/2 at t = 1, plug it in to the general solution to find C.

1/2 = 12/4 - 1/3 + 1/2 + C/12

C = 1/2 - 1/4 + 1/3 - 1/2 = 1/12

So the desired solution is y = t2/4 - t/3 + 1/2 + 1/(12t2).