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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

 Apr 28, 2018
 #1
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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]

 Apr 28, 2018

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