Answer:
Step-by-step 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
