Advertisements
Advertisements
प्रश्न
Solve the differential equation: x dy - y dx = `sqrt(x^2 + y^2)dx,` given that y = 0 when x = 1.
Advertisements
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.
APPEARS IN
संबंधित प्रश्न
Find the particular solution of the differential equation:
2y ex/y dx + (y - 2x ex/y) dy = 0 given that x = 0 when y = 1.
Show that the given differential equation is homogeneous and solve them.
(x2 + xy) dy = (x2 + y2) dx
Show that the given differential equation is homogeneous and solve them.
`{xcos(y/x) + ysin(y/x)}ydx = {ysin (y/x) - xcos(y/x)}xdy`
Show that the given differential equation is homogeneous and solve them.
`x dy/dx - y + x sin (y/x) = 0`
Show that the given differential equation is homogeneous and solve them.
`y dx + x log(y/x)dy - 2x dy = 0`
For the differential equation find a particular solution satisfying the given condition:
(x + y) dy + (x – y) dx = 0; y = 1 when x = 1
Prove that x2 – y2 = c (x2 + y2)2 is the general solution of differential equation (x3 – 3x y2) dx = (y3 – 3x2y) dy, where c is a parameter.
Prove that x2 – y2 = c(x2 + y2)2 is the general solution of the differential equation (x3 – 3xy2)dx = (y3 – 3x2y)dy, where C is parameter
Solve the following initial value problem:
\[x e^{y/x} - y + x\frac{dy}{dx} = 0, y\left( e \right) = 0\]
Solve the following initial value problem:
(xy − y2) dx − x2 dy = 0, y(1) = 1
Solve the following initial value problem:
(y4 − 2x3 y) dx + (x4 − 2xy3) dy = 0, y (1) = 1
Show that the family of curves for which \[\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}\], is given by \[x^2 - y^2 = Cx\]
Which of the following is a homogeneous differential equation?
Solve the following differential equation:
`x * dy/dx - y + x * sin(y/x) = 0`
Solve the following differential equation:
`"y"^2 - "x"^2 "dy"/"dx" = "xy""dy"/"dx"`
Find the equation of a curve passing through `(1, pi/4)` if the slope of the tangent to the curve at any point P(x, y) is `y/x - cos^2 y/x`.
State the type of the differential equation for the equation. xdy – ydx = `sqrt(x^2 + y^2) "d"x` and solve it
F(x, y) = `(sqrt(x^2 + y^2) + y)/x` is a homogeneous function of degree ______.
F(x, y) = `(ycos(y/x) + x)/(xcos(y/x))` is not a homogeneous function.
Solve : `x^2 "dy"/"dx"` = x2 + xy + y2.
Let the solution curve of the differential equation `x (dy)/(dx) - y = sqrt(y^2 + 16x^2)`, y(1) = 3 be y = y(x). Then y(2) is equal to ______.
Find the general solution of the differential equation:
(xy – x2) dy = y2 dx
