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  • Solve for general solution. x dy/dx + 2y = x^2 log x

Solve for General Solution.

    Q6.    x\frac{dy}{dx} + 2y = x^2\log x

Answers (1)

best_answer

Given equation is
x\frac{dy}{dx} + 2y = x^2\log x
Wr can rewrite it as
\frac{dy}{dx} +2.\frac{y}{x}= x\log x
This is  \frac{dy}{dx} + py = Q  type where p = \frac{2}{x} and Q = x\log x
Now,
I.F. = e^{\int pdx}= e^{\int \frac{2}{x}dx}= e^{2\log x}=e^{\log x^2} = x^2                     (\because 0\leq x\leq \frac{\pi}{2} \sec x > 0,\tan x > 0)
Now, the solution of given differential equation is given by relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(x^2) =\int (x\log x\times x^2)dx +C
x^2y = \int x^3\log x+ C
Let  
I = \int x^3\log x\\ \\ I = \log x\int x^3dx-\int \left ( \frac{d(\log x)}{dx}.\int x^3dx \right )dx\\ \\ I = \log x.\frac{x^4}{4}- \int \left ( \frac{1}{x}.\frac{x^4}{4} \right )dx\\ \\ I = \log x.\frac{x^4}{4}- \int \left ( \frac{x^3}{4} \right )dx\\ \\ I = \log x.\frac{x^4}{4}-\frac{x^4}{16}
Put this value in our equation
x^2y =\log x.\frac{x^4}{4}-\frac{x^4}{16}+ C\\ \\ y = \frac{x^2}{16}(4\log x-1)+C.x^{-2}
Therefore, the general solution is  y = \frac{x^2}{16}(4\log x-1)+C.x^{-2}

Posted by

Gautam harsolia

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