Advertisements
Advertisements
प्रश्न
Verify y = `a + b/x` is solution of `x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0
y = `a + b/x`
`(dy)/(dx) = square`
`(d^2y)/(dx^2) = square`
Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`
= `x square + 2 square`
= `square`
Hence y = `a + b/x` is solution of `square`
Advertisements
उत्तर
`x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0
y = `a + b/x`
`("d"y)/("d"x)` = `bb((-b)/x^2)`
`(d^2y)/(dx^2)` = `bb((2b)/x^3)`
Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`
= `bb(x (2b)/x^3 + 2(-2)/x^2)`
= 0
Hence y = `a + b/x` is solution of `bb(x(d^2y)/(dx^2) + 2(dy)/(dx) = 0)`
संबंधित प्रश्न
Show that y = ex (A cos x + B sin x) is the solution of the differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + 2y = 0\]
Verify that y2 = 4a (x + a) is a solution of the differential equations
\[y\left\{ 1 - \left( \frac{dy}{dx} \right)^2 \right\} = 2x\frac{dy}{dx}\]
Differential equation \[\frac{d^2 y}{d x^2} - 3\frac{dy}{dx} + 2y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 3\] Function y = ex + e2x
x cos y dy = (xex log x + ex) dx
tan y \[\frac{dy}{dx}\] = sin (x + y) + sin (x − y)
Solve the following differential equation:
\[\left( 1 + y^2 \right) \tan^{- 1} xdx + 2y\left( 1 + x^2 \right)dy = 0\]
In a bank principal increases at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0.5 = 1.648).
\[\frac{dy}{dx} = \frac{y}{x} + \sin\left( \frac{y}{x} \right)\]
The surface area of a balloon being inflated, changes at a rate proportional to time t. If initially its radius is 1 unit and after 3 seconds it is 2 units, find the radius after time t.
Find the equation of the curve such that the portion of the x-axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2).
At every point on a curve the slope is the sum of the abscissa and the product of the ordinate and the abscissa, and the curve passes through (0, 1). Find the equation of the curve.
Which of the following is the integrating factor of (x log x) \[\frac{dy}{dx} + y\] = 2 log x?
If xmyn = (x + y)m+n, prove that \[\frac{dy}{dx} = \frac{y}{x} .\]
Form the differential equation representing the family of curves y = a sin (x + b), where a, b are arbitrary constant.
Solve the differential equation:
`"x"("dy")/("dx")+"y"=3"x"^2-2`
Find the particular solution of the differential equation `"dy"/"dx" = "xy"/("x"^2+"y"^2),`given that y = 1 when x = 0
Solve the following differential equation y log y = `(log y - x) ("d"y)/("d"x)`
Choose the correct alternative:
Differential equation of the function c + 4yx = 0 is
A solution of differential equation which can be obtained from the general solution by giving particular values to the arbitrary constant is called ______ solution
The function y = cx is the solution of differential equation `("d"y)/("d"x) = y/x`
Find the particular solution of the following differential equation
`("d"y)/("d"x)` = e2y cos x, when x = `pi/6`, y = 0.
Solution: The given D.E. is `("d"y)/("d"x)` = e2y cos x
∴ `1/"e"^(2y) "d"y` = cos x dx
Integrating, we get
`int square "d"y` = cos x dx
∴ `("e"^(-2y))/(-2)` = sin x + c1
∴ e–2y = – 2sin x – 2c1
∴ `square` = c, where c = – 2c1
This is general solution.
When x = `pi/6`, y = 0, we have
`"e"^0 + 2sin pi/6` = c
∴ c = `square`
∴ particular solution is `square`
Solve `x^2 "dy"/"dx" - xy = 1 + cos(y/x)`, x ≠ 0 and x = 1, y = `pi/2`
