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Question:

\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

\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}