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Find the general solution for each differential equation. xlnxdydx+y=2x2x \ln x \frac{d y}{d x}+y=2 x^{2}


Answered 4 months ago
Answered 4 months ago
Step 1
1 of 2

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


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

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

I=eP dx=e1/(xlnx) dx=e1/t dt=elnt=t=lnxI=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 Dx(Iy)D_x(Iy) and integrating the resulting equation.

lnxdydx+1xy=2x{Step 3 : Multiplying each term by I(x)}or, Dx(lnxy)=2x{Step 4 : Replace sum on the left with Dx(I(x)y)}or, lnxy=2x dx+C{Step 5 : Integrate both sides}or, lnxy=x2+Cor, y=x2+Clnx{Step 6 : Solve for y(x)}\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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