Advertisements
Advertisements
प्रश्न
For the differential equation find a particular solution satisfying the given condition:
`2xy + y^2 - 2x^2 dy/dx = 0; y = 2` when x = 1
Advertisements
उत्तर
The given equation
`2xy + y^2 - 2x^2 dy/dx = 0`
or `dy/dx = (2xy + y^2)/(2x^2)`
`= y/x + 1/2(y/x)^2` ....(i)
Clearly, this equation is a differential equation.
∴ putting y = vx
`dy/dx = v + x (dv)/dx` ....(in equation (i))
`v + x (dv)/dx = v + 1/2 v^2`
`=> x (dv)/dx = 1/2 v^2`
`=> 2 xx 1/v^2 (dv) = 1/x dv`
On integrating,
`2 int 1/v^2 dv = int 1/x dv - 2/v = log |x| + C`
So, on substituting `(y/x)` in place of v,
`- (2x)/y = log |x| + C` ....(ii)
Given y = 2 if x = 1 ...[from equation (ii)]
`(- 2)/2 = log1 + C`
- 1 = 0 + C
⇒ C = - 1
On putting C = – 1 in equation (ii)
`(- 2x)/y = log |x| - 1`
y = `(2x)/(1 - log |x|)`
APPEARS IN
संबंधित प्रश्न
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.
`x^2 dy/dx = x^2 - 2y^2 + xy`
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:
(x + y) dy + (x – y) dx = 0; y = 1 when x = 1
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
(x2 − 2xy) dy + (x2 − 3xy + 2y2) dx = 0
(x2 + 3xy + y2) dx − x2 dy = 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:
\[x\frac{dy}{dx} - y + x \sin\left( \frac{y}{x} \right) = 0, y\left( 2 \right) = x\]
Solve the following differential equation:
`"x" sin ("y"/"x") "dy" = ["y" sin ("y"/"x") - "x"] "dx"`
Solve the following differential equation:
`(1 + 2"e"^("x"/"y")) + 2"e"^("x"/"y")(1 - "x"/"y") "dy"/"dx" = 0`
Solve the following differential equation:
`"xy" "dy"/"dx" = "x"^2 + "2y"^2, "y"(1) = 0`
Solve the following differential equation:
(x2 + 3xy + y2)dx - x2 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`.
F(x, y) = `(sqrt(x^2 + y^2) + y)/x` is a homogeneous function of degree ______.
Solve : `x^2 "dy"/"dx"` = x2 + xy + y2.
Solcve: `x ("d"y)/("d"x) = y(log y – log x + 1)`
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 ______.
If a curve y = f(x), passing through the point (1, 2), is the solution of the differential equation, 2x2dy = (2xy + y2)dx, then `f(1/2)` is equal to ______.
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 ______.
A function \[F(x,y)\] is homogeneous of degree \[n\] when which condition holds?
If \[F(\lambda x,\lambda y)=F(x,y)\] for any non-zero constant \[\lambda\], what is the degree of \[F(x,y)\]?
For the substitution \[y=vx\], which differentiated form is correct?
Which dependence is checked on the right-hand side of a homogeneous differential equation?
Which substitutions are chosen for a homogeneous differential equation?
Writing \[x\cos\left(\frac{y}{x}\right)\frac{dy}{dx}=y\cos\left(\frac{y}{x}\right)+x\] in standard form gives which expression?
After putting \[y=vx\] in \[\frac{dy}{dx}=\frac{y\cos\left(\frac{y}{x}\right)+x}{x\cos\left(\frac{y}{x}\right)}\], which equation results?
Which separable equation follows from \[x\frac{dv}{dx}=\frac{1}{\cos v}\]?
