Question

Solve the differential equation: x dy - y dx = `sqrt(x^2 + y^2)dx,` given that y = 0 when x = 1.

Answer 1

xdy - ydx = `sqrt(x^2 + y^2)dx`

⇒ xdy = `[ y + sqrt(x^2+y^2)]dx`

`dy/dx = (y + sqrt(x^2+y^2))/x`       ...(1)

Let F (x,y) = `(y + sqrt(x^2+y^2))/x` 

∴ `"F"(lambdax,lambday) = (lambdax+sqrt((lambdax)^2+ (lambday)^2))/(lambdax) = (y + sqrt(x^2+y^2))/(x) = lambda^0 . "F"(x,y)`

Therefore, the given differential equation is a homogeneous equation.To solve it, we make the substitution as:

y = vx

⇒ `d/dx (y) = d/dx (vx)`
⇒ `dy/dx = v + x (dv)/(dx)`

Substituting the values of v and `dy/dx` in equation (1), we get:

`v + x (dv)/dx = (vx+sqrt(x^2 + (vx)^2))/x`

⇒ `v + x (dv)/dx = v + sqrt(1+v^2)`

⇒ `(dv)/sqrt(1+v^2) = dx/x`

Integrating both sides, we get:

`log |v + sqrt(1+v^2)| = log|x| + log "C"`

⇒ `log |y/x + sqrt(1+y^2/x^2)| = log|"C"x|`

⇒ `log|(y + sqrt(x^2+y^2))/x| = log|"C"x|`

⇒ `y + sqrt(x^2+y^2) = "C"x^2`

This is the required solution of the given differential equation.