MHB Is the given function a solution of the 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
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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