Advertisements
Advertisements
प्रश्न
Solve the following differential equation
x2y dx – (x3 + y3)dy = 0
Advertisements
उत्तर
x2y dx – (x3 + y3)dy = 0
⇒ (x3 + y3).dy = x2y.dx
⇒ `(dy)/(dx) = (x^2y)/(x^3 + y^3)` ......(1)
Let y = vx ......(2)
∴ `(dy)/(dx) = "v" + x (d"v")/(dx)`
∴ Equation (1) can be written as
`"v" + x. (d"v")/(dx) = (x^2."v"x)/(x^3 + "v"^3x^3)`
= `(x^3."v")/(x^3(1 + "v"^3))`
⇒ `"v" + x. (d"v")/(dx) = "v"/(1 + "v"^3)`
⇒ `x. (d"v")/(dx) = "v"/(1 + "v"^3) - "v"`
= `("v" - "v" - "v"^4)/(1 + "v"^3)`
⇒ `x. (d"v")/(dx) = - "v"^4/(1 + "v"^3)`
⇒ `(1 + "v"^3)/"v"^4 . d"v" = - 1/x. dx`
Integrating both sides. we get
`int (1 + "v"^3)/"v"^4. d"v" = - int 1/x. dx`
⇒ `int (1/"v"^4 + 1/"v"). d"v" = - int 1/x. dx`
⇒ `int "v"^-4. d"v" + int 1/"v". d"v" = - int 1/x. dx`
⇒ `"v"^(-3)/(-3) + log |"v"| = - log |x| + C`
⇒ `- 1/3 . 1/"v"^3 + log |"v"| = - log |x| + C`
Resubstituting the value of v from (2),
⇒ `- 1/3 1/(y/x)^3 + log |y/x| = - log |x| + C`
⇒ `(-x^3)/(3y^3) + log |y| - log |x| = - log |x| + C`
∴ `log y = x^3/(3y^3) + C`
APPEARS IN
संबंधित प्रश्न
For the differential equation, find the general solution:
`dy/dx = (1 - cos x)/(1+cos x)`
For the differential equation, find the general solution:
`dy/dx = sqrt(4-y^2) (-2 < y < 2)`
For the differential equation, find the general solution:
`x^5 dy/dx = - y^5`
For the differential equation, find the general solution:
`dy/dx = sin^(-1) x`
For the differential equation, find the general solution:
ex tan y dx + (1 – ex) sec2 y dy = 0
For the differential equation find a particular solution satisfying the given condition:
`(x^3 + x^2 + x + 1) dy/dx = 2x^2 + x; y = 1` When x = 0
For the differential equation find a particular solution satisfying the given condition:
`x(x^2 - 1) dy/dx = 1` , y = 0 when x = 2
For the differential equation find a particular solution satisfying the given condition:
`cos (dx/dy) = a(a in R); y = 1` when x = 0
For the differential equation find a particular solution satisfying the given condition:
`dy/dx` = y tan x; y = 1 when x = 0
Find the equation of a curve passing through the point (0, 0) and whose differential equation is y′ = e x sin x.
For the differential equation `xy(dy)/(dx) = (x + 2)(y + 2)` find the solution curve passing through the point (1, –1).
At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (- 4, -3). Find the equation of the curve given that it passes through (-2, 1).
The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.
In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 doubles itself in 10 years (loge 2 = 0.6931).
The general solution of the differential equation `dy/dx = e^(x+y)` is ______.
Find the particular solution of the differential equation:
`y(1+logx) dx/dy - xlogx = 0`
when y = e2 and x = e
Find the particular solution of the differential equation ex tan y dx + (2 – ex) sec2 y dy = 0, give that `y = pi/4` when x = 0
Verify y = log x + c is a solution of the differential equation
`x(d^2y)/dx^2 + dy/dx = 0`
Solve the differential equation:
`dy/dx = 1 +x+ y + xy`
Solve `dy/dx = (x+y+1)/(x+y-1) when x = 2/3 and y = 1/3`
Solve
y dx – x dy = −log x dx
Solve
`y log y dx/ dy = log y – x`
Solve the differential equation `(x^2 - 1) "dy"/"dx" + 2xy = 1/(x^2 - 1)`.
Solve: (x + y)(dx – dy) = dx + dy. [Hint: Substitute x + y = z after seperating dx and dy]
The solution of the differential equation, `(dy)/(dx)` = (x – y)2, when y (1) = 1, is ______.
