Question:
\frac{d}{dx}\left(x\:\frac{dy}{dx}\:+\left(1+x\right)\:y\right)\:=12\:\:;\:\frac{dy}{dx}=0,\:y=0,\:x=1Solution:
Since the equation is in the form d/dx (dy/dx + y P(x) = Q(x) , we use Case 1
\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.
\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).
\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
\\\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 \left( x \right)= \left(\frac{1}{x}+1\right)
\\Q\left( x \right)= 12-\frac{12}{x}Compute for the integrating factor
\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(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
\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,
\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
\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+\frac{12e^1}{e^x}
\\or
\\yx=12x-24+12e^{1-x}