Question

The function y1(t)=t is a solution to the differential equation t2y′′−t(t+3)y′+(t+3)y=0. ,t > 0; y1(t) = t. Use the method of reduction of order to find a second solution of the given differential equation.

Answer 1

Explanation:

Let y(t) = vt..... 1

Differentiating y(t) with respect to t twice will give;

y'(t) = v(1) + v't

y'(t) = v+v't .... 2

For y''(t);

y''(t) = v'+v'(1)+v''t

y''(t) = 2v'+v''t.... 3

Substitute equation 1, 2 and 3 into the differential equation

t²y′′−t(t+3)y′+(t+3)y=0

t²(2v'+v''t) −t(t+3)(v+v't)+(t+3)(vt)=0

t²(2v'+v''t) −(t²+3t)(v+v't)+(t+3)(vt)=0

Open the parentheses

2t²v'+t³v''-(t²v+t³v'+3tv+3t²v')+t²v+3vt= 0

2t²v'+t³v''-t²v-t³v'-3tv-3t²v'+t²v+3vt=0

Collect like terms according to the degree

t³v''+2t²v'-3t²v'-t³v'-t²v+t²v-3vt+3vt=0

t³v''-t²v'-t³v' =0

t²(tv"-v'-tv') = 0

tv"-v'-tv' = 0

tv"-v'(1+t) = 0

Reduce the degree by substituting p as v' to have;

tp'-p(1+t) = 0

tp' = p(1+t)

Separate the variables

p'/p = (1+t)/t

p'/p = 1/t + 1

Integrate both sides of the equation

ln(p) = lnt + t +K

lnp-lnt = t+K

ln(p/t) = t+K

p/t = e^(t+K)

p = Cte^t

v' = Cte^t

Integrate both sides

v = C{te^t - integral (e^t)}

v(t) = C(te^t - e^t}

y(t)/t = C(te^t - e^t}

y(t) = C(t²e^t-te^t)

Since the solution of second order differential equation

y(t) = C1e^at+C2te^t

In comparison the second solution to the differential equation y(t) will be t²e^t