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
Solve the differential equation (x2 + y2)dx- 2xydy = 0
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.
`y' = (x + y)/x`
Show that the given differential equation is homogeneous and solve them.
(x2 – y2) dx + 2xy dy = 0
Show that the given differential equation is homogeneous and solve them.
`x dy - y dx = sqrt(x^2 + y^2) 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.
`y dx + x log(y/x)dy - 2x dy = 0`
Show that the given differential equation is homogeneous and solve them.
`(1+e^(x/y))dx + e^(x/y) (1 - x/y)dy = 0`
For the differential equation find a particular solution satisfying the given condition:
x2 dy + (xy + y2) dx = 0; y = 1 when x = 1
(x2 − 2xy) dy + (x2 − 3xy + 2y2) dx = 0
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:
\[\frac{dy}{dx} - \frac{y}{x} + cosec\frac{y}{x} = 0, y\left( 1 \right) = 0\]
Solve the following initial value problem:
\[x\frac{dy}{dx} - y + x \sin\left( \frac{y}{x} \right) = 0, y\left( 2 \right) = x\]
Find the particular solution of the differential equation x cos\[\left( \frac{y}{x} \right)\frac{dy}{dx} = y \cos\left( \frac{y}{x} \right) + x\], given that when x = 1, \[y = \frac{\pi}{4}\]
Which of the following is a homogeneous differential equation?
Solve the following differential equation:
`"dy"/"dx" + ("x" - "2y")/("2x" - "y") = 0`
Solve the following differential equation:
`(1 + "e"^("x"/"y"))"dx" + "e"^("x"/"y")(1 - "x"/"y")"dy" = 0`
Solve the following differential equation:
`"y"^2 - "x"^2 "dy"/"dx" = "xy""dy"/"dx"`
Solve the following differential equation:
x dx + 2y dx = 0, when x = 2, y = 1
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
The differential equation y' = `y/(x + sqrt(xy))` has general solution given by:
(where C is a constant of integration)
