Question #193744

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


1
Expert's answer
2021-05-20T07:40:02-0400

D2ydx224dydx+144=0\frac{D^2y}{dx^2}-24\frac{dy}{dx}+144=0


the differential equation is of the form

y+py+qy=sy''+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+(k2+kp)=0q+(k^2+kp)=0

k2+kp=0k^2+kp=0

this is simple equation

the roots of this equation are


k1=0k2=24k_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)=C1ek1x+C2ek2xy(x)=C_1e^{k_1x}+C_2e^{k_2x}

y(x)=C1+C2e24xy(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)=C1(x)+C2(x)e24xy(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:

y1(x)ddxC1(x)+y2(x)ddx2C2(x)=0y_1(x)\frac{d}{dx}C_1(x)+y_2(x)\frac{d}{dx_2}C_2(x)=0\\


ddxC1(x)ddxy1(x)+ddxC2(x)ddxy2(x)=f(x)\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)e^{(24x)} (C1=0 , C2=1)

the free term f=s or

f(x)=144f(x)=-144

so the system has the form


e24xddxC2(x)+ddxC1(x)=0ddx1ddxC1(x)+ddxC2(x)ddxe24x=144ore24xddxC2(x)+ddxC1(x)=024e24xddxC2(x)=144\\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 :

ddxC1(x)=6ddxC2(x)=6e24x\frac{d}{dx}C_1(x)=6 \\ \frac{d}{dx}C_2(x)=-6e^{-24x}

it is the simple equation , solve these equation

C1(x)=C3+6dxC2(x)=C4+(6e24x)dxC_1(x)=C_3+ \int6dx\\ C_2(x)=C_4+\int(-6e^{-24x})dx


or

C1(x)=C3+6xC2(x)=C4+e244C_1(x)=C_3+6x \\ C_2(x)=C_4+\frac{e^{-24}}{4}


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

y(x)=C1(x)+C2(x)e24xy(x)=C_1(x)+C_2(x)e^{24x}


the answer is

y(x)=C3+C4e24x+6x+14y(x)=C_3+C_4e^{24x}+6x+\frac{1}{4}


where C3 and C4 are constant and also14\frac{1}{4}


so final answer is


y(x)=C1+C2e24x+6xanswer\boxed{y{\left(x \right)} = C_{1} + C_{2} e^{24 x} + 6 x}answer




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