हल कीजिए- `x (dy)/(dx) +y-x+xy cot x=0, x ne 0.`

1.	हल कीजिए- `x (dy)/(dx) +y-x+xy cot x=0, x ne 0.`
Answer» दिया गया अवकल समीकरण है- `x(dy)/(dx)+y-xy cot x=0` `impliesx(dy)/(dx) +(1+ x cot x)y=x` `implies (dy)/(dx)+((1+ cot x)/(x))y=1` `implies(dy)/(dx)+ ((1)/(x) +cot x)y=1" "...(1)` यहॉँ `P=1/x+cot x` और `Q=1` `therefore I. F. =e^(int Pdx)=e^(int((1)/(x)+cotx)dx)` `=e^(int(1)/(x)dx+intcot x dx )` `=e ^(log\|x\|+ log \|sin x\|)` `=e ^(log \|x sinx\|)` `=x sin x` अतः अभीष्ट हल है- `yxx (I.F.)=int Qxx (I.F.) dx+C` `impliesy. x sin x=int 1(x sin x)dx +C` `implies xy sin x = int underset(I)(x ) underset(II)(sin) x dx +C` `implies xy sin x=x int sin x dx- int 1xx (-cos x) dx +C` `impliesxy sin x= x (- cos x)+ sin x+C` `implies xy sin x=- x cos x+ sin x+C` `impliesy=(-x cos x)/(x sin x) +(sin x)/(x sin x) +(C)/(x sin x)` `impliesy= -cot x+1/x + (C)/(x sinx).`

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