Advertisements
Advertisements
प्रश्न
Solve the following differential equation:
y2 dx + (xy + x2)dy = 0
Advertisements
उत्तर
y2 dx + (xy + x2)dy = 0
∴ (xy + x2)dy = - y2 dx
∴ `"dy"/"dx" = (- "y"^2)/("xy + x"^2)`
Put y = vx
∴ `"dy"/"dx" = "v + x""dv"/"dx"`
Substituting these values in (1), we get
`"v + x""dv"/"dx" = (- "v"^2"x"^2)/("x" * "vx + x"^2) = (- "v"^2)/("v + 1")`
∴ `"x" "dv"/"dx" = (- "v"^2)/("v + 1") - "v" = (- "v"^2 - "v"^2 - "v")/("v + 1")`
∴ `"x" "dv"/"dx" = (- 2"v"^2 - "v")/("v + 1") = -(("2v"^2 + "v")/("v + 1"))`
∴ `("v + 1")/("2v"^2 + "v")"dv" = - 1/"x" "dx"`
Integrating both sides, we get
`int ("v + 1")/("2v"^2 + "v")"dv" = - int 1/"x" "dx"`
∴`int ("v + 1")/("v"("2v" + 1))"dv" = - int 1/"x" "dx"`
∴ `int(("2v" + 1) - "v")/("v"("2v" + 1))"dv" = - int 1/"x" "dx"`
∴ `int(1/"v" - 1/("2v + 1"))"dv" = - int 1/"x" "dx"`
∴ `int 1/"v" "dv" - int1/("2v + 1")"dv" = - int 1/"x" "dx"`
∴ `log |"v"| - 1/2 log |2"v" + 1| = - log |"x"| + log "c"`
∴ `2 log |"v"| - log |2"v" + 1| = - 2 log |"x"| + 2 log "c"`
∴ `log |"v"^2| - log |"2v" + 1| = - log |"x"^2| + log "c"^2`
∴ `log |"v"^2/("2v" + 1)| = log |"c"^2/"x"^2|`
∴ `"v"^2/("2v" + 1) = "c"^2/"x"^2`
∴ `("y"^2/"x"^2)/(2 ("y"/"x") + 1) = "c"^2/"x"^2`
∴ `"y"^2/("x"("2y + x")) = "c"^2/"x"^2`
∴ xy2 = c2(x + 2y)
This is the general solution.
APPEARS IN
संबंधित प्रश्न
Solve the differential equation (x2 + y2)dx- 2xydy = 0
Show that the differential equation 2yx/y dx + (y − 2x ex/y) dy = 0 is homogeneous. Find the particular solution of this differential equation, given that x = 0 when y = 1.
Show that the differential equation `2xydy/dx=x^2+3y^2` is homogeneous and solve it.
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.
`x dy - y dx = sqrt(x^2 + y^2) dx`
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
For the differential equation find a particular solution satisfying the given condition:
`[xsin^2(y/x - y)] dx + x dy = 0; y = pi/4 "when" x = 1`
Find the particular solution of the differential equation `(x - y) dy/dx = (x + 2y)` given that y = 0 when x = 1.
(x2 − 2xy) dy + (x2 − 3xy + 2y2) dx = 0
Solve the following initial value problem:
(x2 + y2) dx = 2xy dy, y (1) = 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
Solve the following initial value problem:
\[\left\{ x \sin^2 \left( \frac{y}{x} \right) - y \right\}dx + x dy = 0, y\left( 1 \right) = \frac{\pi}{4}\]
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, \[y = \frac{\pi}{4}\]
Find the particular solution of the differential equation \[\left( x - y \right)\frac{dy}{dx} = x + 2y\], given that when x = 1, y = 0.
Show that the family of curves for which \[\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}\], is given by \[x^2 - y^2 = Cx\]
A homogeneous differential equation of the form \[\frac{dx}{dy} = h\left( \frac{x}{y} \right)\] can be solved by making the substitution
Solve the differential equation: ` (dy)/(dx) = (x + y )/ (x - y )`
Solve the differential equation: x dy - y dx = `sqrt(x^2 + y^2)dx,` given that y = 0 when x = 1.
Solve the following differential equation:
`x * dy/dx - y + x * sin(y/x) = 0`
Solve the following differential equation:
`(1 + "e"^("x"/"y"))"dx" + "e"^("x"/"y")(1 - "x"/"y")"dy" = 0`
Solve the following differential equation:
`"xy" "dy"/"dx" = "x"^2 + "2y"^2, "y"(1) = 0`
Solve the following differential equation:
x dx + 2y dx = 0, when x = 2, y = 1
Solve the following differential equation:
`x^2. dy/dx = x^2 + xy + y^2`
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.
The solution of the differential equation `(1 + e^(x/y)) dx + e^(x/y) (1 + x/y) dy` = 0 is
A homogeneous differential equation of the `(dx)/(dy) = h(x/y)` can be solved by making the substitution.
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 ______.
The differential equation y' = `y/(x + sqrt(xy))` has general solution given by:
(where C is a constant of integration)
Read the following passage:
|
An equation involving derivatives of the dependent variable with respect to the independent variables is called a differential equation. A differential equation of the form `dy/dx` = F(x, y) is said to be homogeneous if F(x, y) is a homogeneous function of degree zero, whereas a function F(x, y) is a homogeneous function of degree n if F(λx, λy) = λn F(x, y). To solve a homogeneous differential equation of the type `dy/dx` = F(x, y) = `g(y/x)`, we make the substitution y = vx and then separate the variables. |
Based on the above, answer the following questions:
- Show that (x2 – y2) dx + 2xy dy = 0 is a differential equation of the type `dy/dx = g(y/x)`. (2)
- Solve the above equation to find its general solution. (2)
The solution of the equation `dy/dx = (3x − 4y − 2)/(3x − 4y − 3)` is ______.
