Show that the given differential equation is homogeneous and solve them. (x – y) dy – (x + y) dx = 0 - Mathematics

प्रश्न

Show that the given differential equation is homogeneous and solve them.

(x – y) dy – (x + y) dx = 0

बेरीज

उत्तर

(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 (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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Methods of Solving First Order, First Degree Differential Equations - Homogeneous Differential Equations

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Q 3 Q 2 Q 4

पाठ 9: Differential Equations - Exercise 9.5 [पृष्ठ ४०६]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12

पाठ 9 Differential Equations
Exercise 9.5 | Q 3 | पृष्ठ ४०६

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