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    Q3.    \frac{dy}{dx} + \frac{y}{x} = x^2

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Given equation is
\frac{dy}{dx} + \frac{y}{x} = x^2
This is  \frac{dy}{dx} + py = Q  type where p = \frac{1}{x} and Q = x^2
Now,
I.F. = e^{\int pdx}= e^{\int \frac{1}{x}dx}= e^{\log x}= x
Now, the solution of given differential equation is given by relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(x) =\int (x^2\times x)dx +C
y(x) =\int (x^3)dx +C\\ y.x= \frac{x^4}{4}+C\\
Therefore, the general solution is yx =\frac{x^4}{4}+C???????

Posted by

Gautam harsolia

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