Advertisements
Advertisements
प्रश्न
x2y dx – (x3 + y3) dy = 0
उत्तर
x2y dx – (x3 + y3) dy = 0
∴ x2y dx – (x3 + y3) = dy
∴ `dy/dx = (x^2y)/(x^3 + y^3)` …(i)
Put y = tx …(ii)
Differentiating w.r.t. x, we get
`dy/dx = t + x dt/dx` …(iii)
Substituting (ii) and (iii) in (i), we get
`t + x dt/dx = (x^2 . tx)/(x^3 + t^3 x^3)`
∴ `t + x dt/dx = (x^3.t)/(x^3(1+t^3))`
∴ `x dt/dx = t/(1+t^3) - t`
∴ `x dt/dx = (t-t-t^4)/(1+t^3)`
∴ `x dt/dx = (-t^4)/(1+t^3)`
∴ `(1+t^3)/t^4dt = - dx/x`
Integrating on both sides, we get
`int(1+t^3)/t^4 dt = - int 1/x dx `
∴ `int (1/t^4 + 1/t) dt = - int 1/x dx`
∴ `int t^-4 dt + int 1/t dt = - int1/x dx`
∴ `t^3/-3 + log |t| = - log |x| + c`
∴ `-1/(3t^3)+ log | t | = - log |x| + c`
∴ `- 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
Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.
Show that y = ax3 + bx2 + c is a solution of the differential equation \[\frac{d^3 y}{d x^3} = 6a\].
Verify that y = − x − 1 is a solution of the differential equation (y − x) dy − (y2 − x2) dx = 0.
Differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 2\] Function y = xex + ex
(ey + 1) cos x dx + ey sin x dy = 0
Solve the differential equation \[x\frac{dy}{dx} + \cot y = 0\] given that \[y = \frac{\pi}{4}\], when \[x=\sqrt{2}\]
Write the differential equation representing the family of straight lines y = Cx + 5, where C is an arbitrary constant.
The solution of the differential equation \[\frac{dy}{dx} = \frac{ax + g}{by + f}\] represents a circle when
Find the equation of the plane passing through the point (1, -2, 1) and perpendicular to the line joining the points A(3, 2, 1) and B(1, 4, 2).
Solve the following differential equation.
`y^3 - dy/dx = x dy/dx`
For each of the following differential equations find the particular solution.
`y (1 + logx)dx/dy - x log x = 0`,
when x=e, y = e2.
Solve the following differential equation.
(x2 − y2 ) dx + 2xy dy = 0
Solve the following differential equation.
`dy/dx + y` = 3
Solve the following differential equation.
`(x + a) dy/dx = – y + a`
A solution of a differential equation which can be obtained from the general solution by giving particular values to the arbitrary constants is called ___________ solution.
Solve the following differential equation `("d"y)/("d"x)` = cos(x + y)
Solution: `("d"y)/("d"x)` = cos(x + y) ......(1)
Put `square`
∴ `1 + ("d"y)/("d"x) = "dv"/("d"x)`
∴ `("d"y)/("d"x) = "dv"/("d"x) - 1`
∴ (1) becomes `"dv"/("d"x) - 1` = cos v
∴ `"dv"/("d"x)` = 1 + cos v
∴ `square` dv = dx
Integrating, we get
`int 1/(1 + cos "v") "d"v = int "d"x`
∴ `int 1/(2cos^2 ("v"/2)) "dv" = int "d"x`
∴ `1/2 int square "dv" = int "d"x`
∴ `1/2* (tan("v"/2))/(1/2)` = x + c
∴ `square` = x + c
Solution of `x("d"y)/("d"x) = y + x tan y/x` is `sin(y/x)` = cx
