Advertisements
Advertisements
प्रश्न
Show that the given differential equation is homogeneous and solve them.
`x^2 dy/dx = x^2 - 2y^2 + xy`
उत्तर
Given `x^2 dy/dx = x^2 - 2y^2 + xy`
⇒ `dy/dx = (x^2 - 2y^2 + xy)/x^2`
`= 1 - 2 (y^2/x) + y/x` ....(1)
Since R.H.S. is of the form `g(y/x)`, and so it is a homogeneous function of degree zero.
Therefore equation (1) is a homogeneous differential equation.
∴ put y = vx
⇒ `dy/dx = v.1 + x (dv)/dx`
Substituting these values of y and `dy/dx` in the given equation, we get
`v + x (dv)/dx = 1 - 2 v^2 + v`
⇒ `x (dv)/dx = 1 - 2v^2`
⇒ `1/x dx = 1/ (1 - 2v^2) dv`
On integration, we get
`log |x| = int 1/ (1 - 2v^2) dv + C`
⇒ `log |x| = 1/2 int (dv)/ ((1/ sqrt2)^2 - v^2) + C`
⇒ `log |x| = 1/2 * 1/ (2* (1/sqrt2)) log |((1/sqrt2)+v)/((1/sqrt2) - v)| + C`
⇒ `log |x| = 1/(2sqrt2) log |(1 + sqrt(2v))/(1 - sqrt(2v))| + C`
⇒ `log |x| = 1/ (2 sqrt2) log |(1 + sqrt2* (y/x))/(1 - sqrt2 * (y/x))| + C`
⇒ `log |x| = 1/ (2sqrt2) log | (x + sqrt(2y))/(x - sqrt(2y))| + C`
Where C is an arbitrary constant
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 – y) dy – (x + y) dx = 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:
(x + y) dy + (x – y) 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`
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
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
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.
Prove that x2 – y2 = c(x2 + y2)2 is the general solution of the differential equation (x3 – 3xy2)dx = (y3 – 3x2y)dy, where C is parameter
(x2 + 3xy + y2) dx − x2 dy = 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:
\[x\frac{dy}{dx} - y + x \sin\left( \frac{y}{x} \right) = 0, y\left( 2 \right) = x\]
Find the particular solution of the differential equation \[\left( x - y \right)\frac{dy}{dx} = x + 2y\], given that when x = 1, y = 0.
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 following differential equation:
`(1 + 2"e"^("x"/"y")) + 2"e"^("x"/"y")(1 - "x"/"y") "dy"/"dx" = 0`
Solve the following differential equation:
`"dy"/"dx" + ("x" - "2y")/("2x" - "y") = 0`
Solve the following differential equation:
`x * dy/dx - y + x * sin(y/x) = 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
Solve the following differential equation:
(9x + 5y) dy + (15x + 11y)dx = 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`.
State the type of the differential equation for the equation. xdy – ydx = `sqrt(x^2 + y^2) "d"x` and solve it
Which of the following is not a homogeneous function of x and y.
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.
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 ______.
