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mihalych1998 [28]
3 years ago
11

Use the method of reduction of order to find a second solution to t^2y' + 3ty' – 3y = 0, t> 0 Given yı(t) = t y2(t) = Preview

Give your answer in simplest form (ie no coefficients)
Mathematics
1 answer:
Anestetic [448]3 years ago
6 0

Let y_2(t)=tv(t). Then

{y_2}'=tv'+v

{y_2}''=tv''+2v'

and substituting these into the ODE gives

t^2(tv''+2v')+3t(tv'+v)-3tv=0

t^3v''+5t^2v'=0

tv''+5v'=0

Let u(t)=v'(t), so that u'(t)=v''(t). Then the ODE is linear in u, with

tu'+5u=0

Multiply both sides by t^4, so that the left side can be condensed as the derivative of a product:

t^5u'+5t^4u=(t^5u)'=0

Integrating both sides and solving for u(t) gives

t^5u=C\implies u=Ct^{-5}

Integrate again to solve for v(t):

v=C_1t^{-6}+C_2

and finally, solve for y_2(t) by multiplying both sides by t:

tv=y_2=C_1t^{-5}+C_2t

y_1(t)=t already accounts for the t term in this solution, so the other independent solution is y_2(t)=t^{-5}.

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