MHB Is the given function a solution of the differential equation?

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The discussion confirms that both functions, \( y_1(t) = t/3 \) and \( y_2(t) = e^{-t} + t/3 \), are solutions to the differential equation \( y'''' + 4y''' + 3y = t \). For \( y_1(t) \), the calculations show that substituting it into the equation yields \( t = t \). Similarly, for \( y_2(t) \), the substitution also simplifies correctly to \( t = t \). There are noted typos in the calculations, but the overall verification remains accurate. The conclusion is that both functions satisfy the given differential equation.
karush
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$\textsf{ Verify the following given functions is a solution of the differential equation}\\ \\$
$y''''+4y'''+3y=t\\$
$y_1(t)=t/3$
\begin{align*}
(t/3)''''+4(t/3)'''+(t/3)&=t\\
0+0+t&=t
\end{align*}
$y_2(t)=e^{-t}+t/3$
\begin{align*}
(e^{-t}+t/3)''''+4(e^{-t}+t/3)'''+3(e^{-t}+t/3)&=t\\
e^{-t}-4e^{-t}+3e^{-t}+3e^{-t}+t&=t\\
t&=t
\end{align*}

so is it Raj now?

$$\tiny{\textsf{Elementary Differential Equations and Boundary Value Problems}}$$
 
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A bunch of typos but it's correct. First one it's $3 \cdot (t/3)$, and second one you copied $3e^{-t}$ twice.
 
$\textsf{ Verify the following given functions is a solution of the differential equation}\\ \\$
$y''''+4y'''+3y=t\\$
$y_1(t)=t/3$
\begin{align*}
(t/3)''''+4(t/3)'''+3(t/3)&=t\\
0+0+t&=t
\end{align*}
$y_2(t)=e^{-t}+t/3$
\begin{align*}
(e^{-t}+t/3)''''+4(e^{-t}+t/3)'''+3(e^{-t}+t/3)&=t\\
e^{-t}-4e^{-t}+3e^{-t}+t&=t\\
t&=t
\end{align*}
 
Rido12 said:
A bunch of typos but it's correct. First one it's $3 \cdot (t/3)$, and second one you copied $3e^{-t}$ twice.
mahalo
 

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