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Question
Show that the given differential equation is homogeneous and solve them.
(x – y) dy – (x + y) dx = 0
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Solution
(x - y) dy - (x + y) dx = 0
`=> dy/dx = (x + y)/(x - y)`
`= (1 + (y/x))/(1 - (y/x))`
∵ The powers of the numerator and denominator are the same so this is a homogeneous differential equation.
∴ Putting y = vx
From equation (i),
`dy/dx = v + x (dv)/dx`
`v + x (dv)/dx = (x + vx)/(x - vx)`
`=> x (dv)/dx = (1 + v)/(1 - v) - v`
`=> x dy/dx= (1 + v - v + v^2)/(1 - v)`
`=> x (dv)/dx = (1 + v^2)/(1 - v)`
`=> ((1 - v)/(1 + v^2)) dv = dx/x`
Integrating on both sides
`=> int ((1 - v)/(1 + v^2)) dv = 1/x dx`
`=> int 1/(v^2 + 1) v - 1/2 int (2v)/(v^2 + 1) dv = int 1/x dx`
`=> tan^-1 v = 1/2 log (v^2 + 1) + log x + C`
`=> tan^-1 v = 1/2 log (v^2 + 1) + log x + C`
`=> tan^-1 (y/x) = 1/2 log (y^2/x^2 + 1) + log x + C because y = vx`
`=> tan^-1 (y/x) = 1/2 log ((y^2 + x^2)/x^2) + log x + C`
`=> tan^-1 (y/x) = 1/2 log (x^2 + y^2) - 1/2 log x^2 + log x + C`
`=> tan^-1 (y/x) = 1/2 log (x^2 + y^2) + C`
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An equation involving derivatives of the dependent variable with respect to the independent variables is called a differential equation. A differential equation of the form `dy/dx` = F(x, y) is said to be homogeneous if F(x, y) is a homogeneous function of degree zero, whereas a function F(x, y) is a homogeneous function of degree n if F(λx, λy) = λn F(x, y). To solve a homogeneous differential equation of the type `dy/dx` = F(x, y) = `g(y/x)`, we make the substitution y = vx and then separate the variables. |
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