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