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  • Solve for general solution. xdy/dx + y - x + xy cot x = 0 ( x not equal to 0)

Solve for general solution.

    Q9.    x\frac{dy}{dx} + y -x +xy \cot x = 0\ (x \neq 0)

Answers (1)

best_answer

Given equation is
x\frac{dy}{dx} + y -x +xy \cot x = 0\ (x \neq 0)
we can rewrite it as
\frac{dy}{dx}+y.\left ( \frac{1}{x}+\cot x \right )= 1
This is  \frac{dy}{dx} + py = Q  type where p =\left ( \frac{1}{x}+\cot x \right ) and Q =1
Now,
I.F. = e^{\int pdx}= e^{\int \left ( \frac{1}{x}+\cot x \right ) dx}= e^{\log x +\log |\sin x|} = x.\sin x                     
Now, the solution of the given differential equation is given by the relation
y(I.F.) =\int (Q\times I.F.)dx +C
y(x.\sin x) =\int 1\times x\sin xdx +C
y(x.\sin x) =\int x\sin xdx +C
Lets take
 I=\int x\sin xdx \\ \\ I = x .\int \sin xdx-\int \left ( \frac{d(x)}{dx}.\int \sin xdx \right )dx\\ \\ I =- x.\cos x+ \int (\cos x)dx\\ \\ I = -x\cos x+\sin x
Put this value in our equation
y(x.\sin x)= -x\cos x+\sin x + C\\ y = -\cot x+\frac{1}{x}+\frac{C}{x\sin x}
Therefore, the general solution is   y = -\cot x+\frac{1}{x}+\frac{C}{x\sin x}

Posted by

Gautam harsolia

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