The general solution to differential equation y” + frac2ty' = 0 for t > 0 is given by y(t) = c1

First, we need to find the complementary solution, which is the solution to the homogeneous equa

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The general solution to differential equation $y'' + frac{2}{t}y' = 0$ for $t > 0$ is given by $y(t) = c_1 + frac{c_2}{t}$. Find the general solution to $y'' + frac{2}{t}y' = sin t$. $circ y(t) = c_1 + frac{c_2}{t} + int t^2 sin t , dt - frac{1}{t} int t sin t , dt$ $circ y(t) = c_1 + frac{c_2}{t} + int t sin t , dt - frac{1}{t} int t^2 sin t , dt$ $circ y(t) = c_1 + frac{c_2}{t} + int frac{-sin t}{t ln t} , dt + frac{1}{t} int frac{sin t}{ln t} , dt$ $circ y(t) = c_1 + frac{c_2}{t} + int frac{-sin t}{t^3} , dt + int frac{sin t}{t^2} , dt$

          The general solution to differential equation $y'' + frac{2}{t}y' = 0$ for $t > 0$ is given by $y(t) = c_1 + frac{c_2}{t}$.
Find the general solution to $y'' + frac{2}{t}y' = sin t$.
$circ y(t) = c_1 + frac{c_2}{t} + int t^2 sin t , dt - frac{1}{t} int t sin t , dt$
$circ y(t) = c_1 + frac{c_2}{t} + int t sin t , dt - frac{1}{t} int t^2 sin t , dt$
$circ y(t) = c_1 + frac{c_2}{t} + int frac{-sin t}{t ln t} , dt + frac{1}{t} int frac{sin t}{ln t} , dt$
$circ y(t) = c_1 + frac{c_2}{t} + int frac{-sin t}{t^3} , dt + int frac{sin t}{t^2} , dt$

$The general solution to differential equation y” + frac2ty' = 0 for t > 0 is given by y(t) = c1 + fracc2t. Find the general solution to y” + frac2ty' = sin t. circ y(t) = c1 + fracc2t + int t^2 sin t , dt - frac1t int t sin t , dt circ y(t) = c1 + fracc2t + int t sin t , dt - frac1t int t^2 sin t , dt circ y(t) = c1 + fracc2t + int frac-sin tt ln t , dt + frac1t int fracsin tln t , dt circ y(t) = c1 + fracc2t + int frac-sin tt^3 , dt + int fracsin tt^2 , dt$

Added by Mia H.

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