-1.3.9 Verify ty'-y=t^2 is a solution of the DE

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The discussion verifies that the function \( y_1(t) = 3t + t^2 \) is a solution to the differential equation \( ty' - y = t^2 \). The verification process involves differentiating \( y_1(t) \), substituting it back into the equation, and confirming that both sides equal \( t^2 \). The calculations show that \( t(3 + 2t) - (3t + t^2) = t^2 \), establishing the solution definitively. The participants note the simplicity of the problem, attributing it to foundational knowledge in Calculus.

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karush
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$\textsf{ Verify the following given functions is a solution of the differential equation}\\ \\$
$ty'-y=t^2\\$
$y_1(t)=3t+t^2$
\begin{align*}
t(3t+t^2)'-(3t+t^2)&=t^2\\
t(3+2t)-(3t+t^2)&=\\
3t+2t^2-3t-t^2&=\\
t^2&=t^2
\end{align*}

probably too easy
 
Last edited:
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Well, learning Calculus does make some problems easy!
 

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