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Elementary Differential Equations by Dela Fuente, Feliciano, and Uy Chapter 9 Problem 2 — Special Second-Ordered Differential Equations

Question:

ddx(x dydx +(1+x) y) =12  ; dydx=0, y=0, x=1\frac{d}{dx}\left(x\:\frac{dy}{dx}\:+\left(1+x\right)\:y\right)\:=12\:\:;\:\frac{dy}{dx}=0,\:y=0,\:x=1

Solution:

Since the equation is in the form d/dx (dy/dx + y P(x) = Q(x) , we use Case 1

let u=xdydx+(1+x) y∫ dudx =∫ 12u=12x+C1\begin{align*} let\:u & = x\frac{dy}{dx}+\left(1+x\right)\:y \\ \int \:\frac{du}{dx}\:&=\int \:12 \\u &=12x+C_1 \end{align*}

Solving for the value of C1 using the initial values.

xdydx+(1+x)y =12x +C1 ; dydx=0, y=0, x=11(0)+(1+1)0=12(1)+C10+0=12+C1−12=C1\begin{align*} x\frac{dy}{dx}+\left(1+x\right)y\: & =12x\:+C_1\:;\:\frac{dy}{dx}=0,\:y=0,\:x=1 \\1(0) + (1+1)0 & = 12(1) + C_1 \\0+0 & =12+C_1 \\-12 & =C_1 \end{align*}

Rewriting the equation into the general form of a first-order linear differential equation (FOLDE).

[xdydx+(1+x)y=12x−12]1xdydx+(1+xx)y=12−12xdydx+(1x+1)y=12−12x\left[x\frac{dy}{dx}+\left(1+x\right)y=12x-12\right]\frac{1}{x} \\\frac{dy}{dx}+\left(\frac{1+x}{x}\right)y=12-\frac{12}{x} \\\frac{dy}{dx}+\left(\frac{1}{x}+1\right)y=12-\frac{12}{x}

Since the equation is now in the form of dy/dx + y P(x) = Q(x), we use FOLDE

dydx+(1x+1)y=12−12x\\\frac{dy}{dx}+\left(\frac{1}{x}+1\right)y=12-\frac{12}{x}

From the general form of a first-order differential equation, we have

P(x)=(1x+1)Q(x)=12−12x\\P \left( x \right)= \left(\frac{1}{x}+1\right) \\Q\left( x \right)= 12-\frac{12}{x}

Compute for the integrating factor

ϕ=e∫P(x)dxϕ=e∫ (1x+1)dxϕ =eln⁡x+xϕ =x(ex)\begin{align*} \phi &= e^ {\int P\left(x \right) dx} \\\phi & =e^{\int \:\left(\frac{1}{x}+1\right)dx} \\\phi & \:=e^{\ln x+x} \\\phi & \:=x\left(e^x\right) \end{align*}

Substituting everything to the solution of a first-order linear differential equation, we have

y(xex)=∫xex(12−12x)dx+C2y(xex)=∫(12xex−12ex)dx+C2yxex=∫12xex−∫ 12ex dx+C2​y(xe^x)=\int xe^x\left(12-\frac{12}{x}\right)dx+C_2 \\y\left(xe^x\right)=\int \left(12xe^x-12e^x\right)dx+C_2 \\yxe^x=\int 12xe^x-\int \:12e^x\:dx+C_2​

Use Integration by Parts to solve for the first integral

∫ 12xexdx=12∫xexdxu=xdu=dxdv=ex v=exThereforeuv−∫ vdu=xex−∫ exdx=xex−exConsequently∫ 12xexdx=12(xex−ex)\begin{align*} \int \:12xe^xdx & = 12 \int xe^x dx\\ u = x & &du=dx \\ dv = & e^x \ & v=e^x \\ \text{Therefore} \\ uv-\int \:vdu & =xe^x-\int \:e^xdx \\ & =xe^x-e^x \\ \text{Consequently} \\ \int \:12xe^xdx & = 12 \left( xe^x-e^x\right) \end{align*}

Therefore,

yxex=12(xex−ex)−12ex+C2yxex=12xex−12ex−12ex+C2yxex=12xex−24ex+C2\begin{align*} yxe^x & =12\left(xe^x-e^x\right)-12e^x+C_2 \\yxe^x &=12xe^x-12e^x-12e^x+C_2 \\yxe^x &=12xe^x-24e^x+C_2 \end{align*}

Solving for C2

yx=12x−24+C2ex  ; y=0, x=10(1)=12(1)−24+C2e1 0=12−24e+C2e1 0=−12+C2e1 12e1=C2 \begin{align*} yx & =12x-24+\frac{C_2}{e^x}\:\:;\:y=0,\:x=1 \\0\left(1\right) & =12\left(1\right)-24+\frac{C_2}{e^1}\: \\0 & =12-24e+\frac{C_2}{e^1}\: \\0 &=-12+\frac{C_2}{e^1}\: \\12e^1 & =C_2\: \end{align*}

Therefore, the solution to the problem is

yx=12x−24+12e1exoryx=12x−24+12e1−xyx=12x-24+\frac{12e^1}{e^x} \\or \\yx=12x-24+12e^{1-x}