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प्रश्न
Solve the following differential equation.
y2 dx + (xy + x2 ) dy = 0
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उत्तर
y2 dx + (xy + x2 ) dy = 0
∴ (xy + x2 ) dy = - y2 dx
∴`dy/dx = (-y^2)/(xy+x^2)` ...(i)
Put y = tx ...(ii)
Differentiating w.r.t. x, we get
`dy/dx = t + x dt/dx` ...(iii)
Substituting (ii) and (iii) in (i), we get
`t + x dt/dx = (-t^2x^2)/(x.tx+x^2)`
∴`t + x dt/dx = (-t^2x^2)/(tx^2+x^2)`
∴ `t + x dt/dx = (-t^2x^2)/(x^2(t+1)`
∴ ` x dt/dx = (-t^2)/(t+1)-t`
∴ ` x dt/dx = (-t^2-t^2-t)/(t+1)`
∴ ` x dt/dx = (-(2t^2+t))/(t+1)`
∴ `(t+1)/(2t^2+t)dt = - 1/x dx`
Integrating on both sides, we get
`int (t+1)/(2t^2+t)dt = -int1/xdx`
∴`int (2t + 1 - t)/(t(2t+1)) dt = - int1/xdx`
∴`int1/tdt-int 1/ (2t+1) dt = -int1/ x dx`
∴ log | t | - `1/2` log |2t + 1| = - log |x| + log |c|
∴ 2log| t | - log |2t + 1| = - 2log |x| + 2 log |c|
∴ `2log |y/x | -log |(2y)/x+ 1 |= - 2log |x| + 2 log |c|`
∴ 2log |y| - 2log |x| - log |2y + x| + log |x|
= -2log |x| + 2log |c|
∴ log |y2 | + log |x| = log |c2 | + log |2y + x|
∴ log |y2 x| = log | c2 (x + 2y)|
∴ xy 2 = c2 (x + 2y)
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