Advertisements
Advertisements
प्रश्न
Solve the following differential equation:
`"xy" "dy"/"dx" = "x"^2 + "2y"^2, "y"(1) = 0`
Advertisements
उत्तर
`"xy" "dy"/"dx" = "x"^2 + "2y"^2, "y"(1) = 0`
∴ `"dy"/"dx" = ("x"^2 + "2y"^2)/"xy"` ....(1)
Put y = vx. Then `"dy"/"dx" = "v + x" "dv"/"dx"`
∴ (1) becomes, `"v + x" "dv"/"dx" = ("x"^2 + 2"v"^2"x"^2)/("x" * "vx") = (1 + "2v"^2)/"v"`
∴ `"x" "dv"/"dx" = (1 + "2v"^2)/"v" - "v" = (1 + 2"v"^2 - "v"^2)/"v"`
∴ `"x" "dv"/"dx" = (1 + "v"^2)/"v"`
∴ `"v"/(1 + "v"^2) "dv" = 1/"x" "dx"`
Integrating, we get
∴ `int "v"/(1 + "v"^2) "dv" = int 1/"x" "dx"`
∴ `1/2 int "2v"/(1 + "v"^2) "dv" = int1/"x" "dx" + log "c"_1`
∴ `1/2 log |1 + "v"^2| = log |"x"| + log "c"_1`
∴ `log |1 + "v"^2| = 2 log |"x"^2| + 2 log "c"_1^2`
∴ `log |1 + "v"^2| = log |"cx"^2|, "where" "c" = "c"_1^2`
∴ 1 + v2 = cx2
∴ `1 + "y"^2/"x"^2 = "cx"^2`
∴ `("x"^2 + "y"^2)/"x"^2 = "cx"^2`
∴ `"x"^2 + "y"^2 = "cx"^4`
This is the general solution.
Now, y(1) = 0, i.e. when x = 1, y = 0, we get
1 + 0 = c(1)
∴ c = 1
∴ the particular solution is `"x"^2 + "y"^2 = "x"^4`.
Notes
The answer in the textbook is incorrect.
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.
Solve the differential equation :
`y+x dy/dx=x−y dy/dx`
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^2 dy/dx = x^2 - 2y^2 + xy`
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`
For the differential equation find a particular solution satisfying the given condition:
`dy/dx - y/x + cosec (y/x) = 0; y = 0` when x = 1
A homogeneous differential equation of the from `dx/dy = h (x/y)` can be solved by making the substitution.
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
(2x2 y + y3) dx + (xy2 − 3x3) dy = 0
Solve the following initial value problem:
x (x2 + 3y2) dx + y (y2 + 3x2) dy = 0, y (1) = 1
Solve the following initial value problem:
\[\left\{ x \sin^2 \left( \frac{y}{x} \right) - y \right\}dx + x dy = 0, y\left( 1 \right) = \frac{\pi}{4}\]
Solve the following initial value problem:
\[x\frac{dy}{dx} - y + x \sin\left( \frac{y}{x} \right) = 0, y\left( 2 \right) = x\]
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:
y2 dx + (xy + x2)dy = 0
Solve the following differential equation:
`x * dy/dx - y + x * sin(y/x) = 0`
Solve the following differential equation:
x dx + 2y dx = 0, when x = 2, y = 1
Solve the following differential equation:
(x2 + 3xy + y2)dx - x2 dy = 0
Solve the following differential equation:
(x2 – y2)dx + 2xy dy = 0
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) = `(x^2 + y^2)/(x - y)` is a homogeneous function of degree 1.
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
