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 :
`y+x dy/dx=x−y dy/dx`
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.
`x dy - y dx = sqrt(x^2 + y^2) dx`
Show that the given differential equation is homogeneous and solve them.
`x dy/dx - y + x sin (y/x) = 0`
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`
Which of the following is a homogeneous differential equation?
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.
Find the particular solution of the differential equation `(x - y) dy/dx = (x + 2y)` given that y = 0 when x = 1.
(x2 + 3xy + y2) dx − x2 dy = 0
Solve the following initial value problem:
(y4 − 2x3 y) dx + (x4 − 2xy3) dy = 0, y (1) = 1
Solve the following initial value problem:
\[\left\{ x \sin^2 \left( \frac{y}{x} \right) - y \right\}dx + x dy = 0, y\left( 1 \right) = \frac{\pi}{4}\]
A homogeneous differential equation of the form \[\frac{dx}{dy} = h\left( \frac{x}{y} \right)\] can be solved by making the substitution
Which of the following is a homogeneous differential equation?
Solve the following differential equation : \[\left[ y - x \cos\left( \frac{y}{x} \right) \right]dy + \left[ y \cos\left( \frac{y}{x} \right) - 2x \sin\left( \frac{y}{x} \right) \right]dx = 0\] .
Solve the differential equation: ` (dy)/(dx) = (x + y )/ (x - y )`
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:
y2 dx + (xy + x2)dy = 0
Solve the following differential equation:
`x^2. dy/dx = x^2 + xy + y^2`
Solve the following differential equation:
(9x + 5y) dy + (15x + 11y)dx = 0
F(x, y) = `(x^2 + y^2)/(x - y)` is a homogeneous function of degree 1.
Solve : `x^2 "dy"/"dx"` = x2 + xy + y2.
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)
Which dependence is checked on the right-hand side of a homogeneous differential equation?
What form should the equation be reduced to before integration?
For \[F(x,y)=\frac{y\cos\left(\frac{y}{x}\right)+x}{x\cos\left(\frac{y}{x}\right)}\], what is \[F(\lambda x,\lambda y)\]?
