+9 t(x')(t)+x(t)=t^2
I have the following problem and i know the answer is x(t)=(t^2/3)+(C/t)
I don't know the steps.
An explanation would be greatly appreciated
Solve t ( d^2 x(t))/( dt^2) + 2 ( dx(t))/( dt) = 2 t:
Let ( dx(t))/( dt) = v(t), which gives ( d^2 x(t))/( dt^2) = ( dv(t))/( dt):
Divide both sides by t:
( dv(t))/( dt) + (2 v(t))/t = 2
Let μ(t) = e^( integral2/t dt) = t^2.
Multiply both sides by μ(t):
t^2 ( dv(t))/( dt) + (2 t) v(t) = 2 t^2
Substitute 2 t = d/( dt)(t^2):
t^2 ( dv(t))/( dt) + d/( dt)(t^2) v(t) = 2 t^2
Apply the reverse product rule f ( dg)/( dt) + g ( df)/( dt) = d/( dt)(f g) to the left-hand side:
d/( dt)(t^2 v(t)) = 2 t^2
Integrate both sides with respect to t:
integral d/( dt)(t^2 v(t)) dt = integral2 t^2 dt
Evaluate the integrals:
t^2 v(t) = (2 t^3)/3 + c_1, where c_1 is an arbitrary constant.
Divide both sides by μ(t) = t^2:
v(t) = (2 t)/3 + c_1/t^2
Substitute back for ( dx(t))/( dt) = v(t):
( dx(t))/( dt) = (2 t)/3 + c_1/t^2
Integrate both sides with respect to t:
x(t) = integral((2 t)/3 + c_1/t^2) dt = t^2/3 - c_1/t + c_2, where c_2 is an arbitrary constant.
Simplify the arbitrary constants:
x(t) = t^2/3 + C1/t + C2. [Courtesy of Mathematica 11 Home Edition]
