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प्रश्न
Show that the given differential equation is homogeneous and solve them.
`x^2 dy/dx = x^2 - 2y^2 + xy`
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उत्तर
Given `x^2 dy/dx = x^2 - 2y^2 + xy`
⇒ `dy/dx = (x^2 - 2y^2 + xy)/x^2`
`= 1 - 2 (y^2/x) + y/x` ....(1)
Since R.H.S. is of the form `g(y/x)`, and so it is a homogeneous function of degree zero.
Therefore equation (1) is a homogeneous differential equation.
∴ put y = vx
⇒ `dy/dx = v.1 + x (dv)/dx`
Substituting these values of y and `dy/dx` in the given equation, we get
`v + x (dv)/dx = 1 - 2 v^2 + v`
⇒ `x (dv)/dx = 1 - 2v^2`
⇒ `1/x dx = 1/ (1 - 2v^2) dv`
On integration, we get
`log |x| = int 1/ (1 - 2v^2) dv + C`
⇒ `log |x| = 1/2 int (dv)/ ((1/ sqrt2)^2 - v^2) + C`
⇒ `log |x| = 1/2 * 1/ (2* (1/sqrt2)) log |((1/sqrt2)+v)/((1/sqrt2) - v)| + C`
⇒ `log |x| = 1/(2sqrt2) log |(1 + sqrt(2v))/(1 - sqrt(2v))| + C`
⇒ `log |x| = 1/ (2 sqrt2) log |(1 + sqrt2* (y/x))/(1 - sqrt2 * (y/x))| + C`
⇒ `log |x| = 1/ (2sqrt2) log | (x + sqrt(2y))/(x - sqrt(2y))| + C`
Where C is an arbitrary constant
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