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Question

# Find the general solution for each differential equation. $x \ln x \frac{d y}{d x}+y=2 x^{2}$

Solution

Verified
Step 1
1 of 2

$\text{\underline{Step 1}}$ : The given differential equation is not in the standard Linear First-order differential equation form :

$\dfrac{dy}{dx}+P(x)y=Q(x)$

To transform the given equation into this form, we divide the entire equation by $x\ln x$ to make the coefficient of $\dfrac{dy}{dx}$ unity giving us $P(x)=\dfrac{1}{x\ln x}$ and $Q(x)=\dfrac{2x}{\ln x}$.

$\text{\underline{Step 2}}$ : Evaluating the integrating factor:

$I=e^{\smallint P\ dx}=e^{\smallint 1/(x\ln x)\ dx}=e^{\smallint 1/t\ dt}=e^{\ln t}=t=\color{#c34632}\ln x$.

Multiplying both sides of the differential equation with the integrating factor, rewriting the left hand side of the equation as $D_x(Iy)$ and integrating the resulting equation.

\begin{align*} \ln x\dfrac{dy}{dx}+\dfrac{1}{x}y&=2x\hspace{20pt}\{\text{\color{#4257b2}Step 3 : Multiplying each term by I(x)}\}\\ \text{or, }D_x(\ln x y)&=2x\hspace{20pt}\{\text{\color{#4257b2}Step 4 : Replace sum on the left with D_x(I(x)y)}\}\\ \text{or, }\ln x y&=\int 2x \ dx+C\hspace{20pt}\{\text{\color{#4257b2}Step 5 : Integrate both sides}\}\\ \text{or, }\ln x y&=x^2+C\\ \text{or, }\color{#c34632}y&\color{#c34632}=\dfrac{x^2+C}{\ln x}\color{default}\hspace{20pt}\{\text{\color{#4257b2}Step 6 : Solve for y(x)}\}\\ \end{align*}

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