Question

Consider the system of differential equations dxdt=−5ydydt=−5x. Convert this system to a second order differential equation in y by differentiating the second equation with respect to t and substituting for x from the first equation. Solve the equation you obtained for y as a function of t; hence find x as a function of t. If we also require x(0)=1 and y(0)=3, what are x and y? x(t)= equation editorEquation Editor y(t)= equation editorEquation Editor

Answer 1

$\dfrac{\mathrm dx}{\mathrm dt}=-5y$

$\dfrac{\mathrm dy}{\mathrm dt}=-5x\implies\dfrac{\mathrm d^2y}{\mathrm dt^2}=-5\dfrac{\mathrm dx}{\mathrm dt}$

$\implies\dfrac{\mathrm d^2y}{\mathrm dt^2}-25y=0$

This ODE is linear in $y(t)$ with the characteristic equation and roots

$r^2-25=0\implies r=\pm5$

so that

$y(t)=C_1e^{5t}+C_2e^{-5t}$

Then

$\dfrac{\mathrm dx}{\mathrm dt}=-5C_1e^{5t}-5C_2e^{-5t}$

$\implies x(t)=-C_1e^{5t}+C_2e^{-5t}$

Given that $x(0)=1$ and $y(0)=3$, we find

$\begin{cases}1=-C_1+C_2\\3=C_1+C_2\end{cases}\implies C_1=1,C_2=2$

and the particular solution to this system is

$\begin{cases}x(t)=-e^{5t}+2e^{-5t}\\y(t)=e^{5t}+2e^{-5t}\end{cases}$

Answer 2

The value of x and y would be $x(t) =-e^{5t}+ 2e^{-5t}$ and $y(t) = e^{5t}+ 2e^{-5t}$.

What is a differential equation?

An equation containing derivatives of a variable with respect to some other variable quantity is called differential equations.

The derivatives might be of any order, some terms might contain product of derivatives and the variable itself, or with derivatives themselves. They can also be for multiple variables.

we have the following differential equations

$\dfrac{dx}{dt} =-5y\\\\\dfrac{dy}{dt}=-5x$

by differentiating the second equation we have

$\dfrac{d^2y}{dt^2}=-5\dfrac{dx}{dt}\\\\\dfrac{d^2y}{dt^2}-25y = 0$

$r^2 - 25 = 0\\\\r = \pm5$

$y(t) = C_1e^{5t}+ C_2e^{-5t}$

and by using the characteristic polynomial

$x(t) =-e^{5t}+ 2e^{-5t}\\\\y(t) = e^{5t}+ 2e^{-5t}$

Learn more about differential equations;

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