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Daniel [21]
3 years ago
10

Solve the initial-value problem. x' + 2tx = 5t, x(0) = 8 x(t) =

Mathematics
1 answer:
11111nata11111 [884]3 years ago
8 0

Multiply both sides of the ODE

x'+2tx+5t

by e^{t^2}:

e^{t^2}x'+2te^{t^2}x=5te^{t^2}

Now the left side can be condensed as the derivative of a product:

\left(e^{t^2}x\right)'=5te^{t^2}

Integrate both sides, then solve for <em>x</em> :

e^{t^2}x=\dfrac52e^{t^2}+C

\implies x(t)=\dfrac52+Ce^{-t^2}

Given that <em>x</em>(0) = 8, we find

8=\dfrac52+Ce^0\implies C=\dfrac{11}2

so that the particular solution to this IVP is

\boxed{x(t)=\dfrac{5+11e^{-t^2}}2}

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