Answer to Question #193744 in Differential Equations for Talim

Question

D2y/dx2-24dy/dx+144=0

Answer 1

$\frac{D^2y}{dx^2}-24\frac{dy}{dx}+144=0$

the differential equation is of the form

$y''+py'+qy=s$

p=-24

q=0

s=-144

It is linear inhomogeneous

second -order differential equation with constant coefficients.

the equation has an easy solution

first we should find the roots of the characteristic equation

$q+(k^2+kp)=0$

$k^2+kp=0$

this is simple equation

the roots of this equation are

$k_1=0 \\ k_2=24$

as there are two roots of characteristic equation ,

and the roots are not complex , then

solving the corresponding differential equation looks as follows

$\implies$

$y(x)=C_1e^{k_1x}+C_2e^{k_2x}$

$y(x)=C_1+C_2e^{24x}$

we get asolution for the corresponding homogeneous equation

Now we should solve the inhomogeneous equation

y''+py'+qy=s

use variation of parameters method

Suppose that C₁ and C₂ - it is function of x

tjhe general solution is

$y(x)=C_1(x)+C_2(x)e^{24x}$

where C1(x) and C2(x)

by the method of variation of parameters , we find the solution from the system:

$y_1(x)\frac{d}{dx}C_1(x)+y_2(x)\frac{d}{dx_2}C_2(x)=0\\$

$\frac{d}{dx}C_1(x)\frac{d}{dx}y_1(x)+\frac{d}{dx}C_2(x)\frac{d}{dx}y_2(x)=f(x)$

where

y1(x) and y2(x) - linearly independent particular solutions of Linear Ordinary Differential Equations,

y1(x)=1 (C1=1 , C2=0),

y2(x)=$e^{(24x)}$ (C1=0 , C2=1)

the free term f=s or

$f(x)=-144$

so the system has the form

$\\e^{24x}\frac{d}{dx}C_2(x)+\frac{d}{dx}C_1(x)=0\\ \frac{d}{dx}1\frac{d}{dx}C_1(x)+\frac{d}{dx}C_2(x)\frac{d}{dx}e^{24x}=-144 \\ or \\ e^{24x}\frac{d}{dx}C_2(x)+\frac{d}{dx}C_1(x)=0 \\ 24e^{24x}\frac{d}{dx}C_2(x)=-144$

solve the system :

$\frac{d}{dx}C_1(x)=6 \\ \frac{d}{dx}C_2(x)=-6e^{-24x}$

it is the simple equation , solve these equation

$C_1(x)=C_3+ \int6dx\\ C_2(x)=C_4+\int(-6e^{-24x})dx$

or

$C_1(x)=C_3+6x \\ C_2(x)=C_4+\frac{e^{-24}}{4}$

substitute found C1(x) and C2(x) to

$y(x)=C_1(x)+C_2(x)e^{24x}$

the answer is

$y(x)=C_3+C_4e^{24x}+6x+\frac{1}{4}$

where C₃ and C₄ are constant and also$\frac{1}{4}$

so final answer is

$\boxed{y{\left(x \right)} = C_{1} + C_{2} e^{24 x} + 6 x}answer$