MHB Solutions of DE System & 2nd Order Differential Equation

WMDhamnekar
MHB
Messages
376
Reaction score
28
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?
 
Physics news on Phys.org
Dhamnekar Winod said:
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.
 
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?
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top