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  • Find the general solution. x log x dy/dx + y = 2/x log x

Solve for general solutions.

    Q7.    x\log x\frac{dy}{dx} + y = \frac{2}{x}\log x

Answers (1)

best_answer

Given equation is
x\log x\frac{dy}{dx} + y = \frac{2}{x}\log x
we can rewrite it as
\frac{dy}{dx}+\frac{y}{x\log x}= \frac{2}{x^2}
This is  \frac{dy}{dx} + py = Q  type where p = \frac{1}{x\log x} and Q =\frac{2}{x^2}
Now,
I.F. = e^{\int pdx}= e^{\int \frac{1}{x\log x} dx}= e^{\log(\log x)} = \log x                     
Now, the solution of given differential equation is given by relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(\log x) =\int ((\frac{2}{x^2})\times \log x)dx +C

take
I=\int ((\frac{2}{x^2})\times \log x)dx
I = \log x.\int \frac{2}{x^2}dx-\int \left ( \frac{d(\log x)}{dt}.\int \frac{x^2}{2}dx \right )dx \\ \\ I= -\log x . \frac{2}{x}+ \int (\frac{1}{x}.\frac{2}{x})dx\\ \\ I = -\log x.\frac{2}{x}+ \int \frac{2}{x^2}dx\\ \\I = -\log x.\frac{2}{x}- \frac{2}{x}\\ \\
Put this value in our equation

y\log x =-\frac{2}{x}(\log x+1)+C\\ \\
Therefore, the general solution is   y\log x =-\frac{2}{x}(\log x+1)+C\\ \\

Posted by

Gautam harsolia

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