Solutions of DE System & 2nd Order Differential Equation

[WMDhamnekar]

Hello,
$\vec{x'}=\small\begin{pmatrix}1&2\\3&2\end{pmatrix}\vec{x}+t\small\begin{pmatrix}2\\-4\end{pmatrix}$

Now i got the solution to this differential equation system as

$\vec{x}(t)=c_1e^{-t}\small\begin{pmatrix}-1\\1\end{pmatrix}$+$c_2e^{4t}\small\begin{pmatrix}2\\3\end{pmatrix}$+$t\small\begin{pmatrix}3\\\frac{-5}{2}\end{pmatrix}$+$\small\begin{pmatrix}-2.75\\2.875\end{pmatrix}$

Now i converted this differential equation system into ordinary differential equation $y''-3y'-4y+12t-2=0$

I got solution to this DE as $y=C_1e^{-x}+C_2e^{4x}+3t-\frac12$.

Now my question why there is diference in these two solutions?

 

[I like Serena]

Hello,
$\vec{x'}=\small\begin{pmatrix}1&2\\3&2\end{pmatrix}\vec{x}+t\small\begin{pmatrix}2\\-4\end{pmatrix}$

Now i got the solution to this differential equation system as

$\vec{x}(t)=c_1e^{-t}\small\begin{pmatrix}-1\\1\end{pmatrix}$+$c_2e^{4t}\small\begin{pmatrix}2\\3\end{pmatrix}$+$t\small\begin{pmatrix}3\\\frac{-5}{2}\end{pmatrix}$+$\small\begin{pmatrix}-2.75\\2.875\end{pmatrix}$

Now i converted this differential equation system into ordinary differential equation $y''-3y'-4y+12t-2=0$

I got solution to this DE as $y=C_1e^{-x}+C_2e^{4x}+3t-\frac12$.

Now my question why there is diference in these two solutions?

Hi Dhamnekar Winod, welcome to MHB! ;)

Your solution contains $x$ on the right hand side. I presume it should be $t$?
Either way, if I feed the latter equation to Wolfram|Alpha, I get:
$$y(t)=c_1e^{-t}+c_2e^{4t}+3t-\frac{11}{4}$$
So:

• It seems to be the solution for $x(t)$ rather than $y(t)$.
• Your equation is correct, although for $x(t)$ rather than $y(t)$.
• There is a minor arithmetic mistake somewhere in your solution.

 

[WMDhamnekar]

Hello, The derivative of $x_1=C_1e^{-t}+C_23e^{4t}-2.5t+2.875$in the differential equation system.But in the second case,it is different Why?