Advertisements
Advertisements
प्रश्न
Solve
`dy/dx + 2/ x y = x^2`
उत्तर
`dy/dx + 2/ x y = x^2`
The given equation is of the form
`dy/dx + py = Q`
`where, P = 2/x and Q = x^2`
∴ I.F. =`e^(int^(pdx) = e^(2int^(1/xdx) e = ^(2logx) = e^(logx^2) = x^2`
∴ Solution of the given equation is
`y(I.F.) = int Q(I.F.) dx + c_1`
`y(x^2) = int x^2 xx x^2 dx + c_1`
∴ `x ^2 y = x^4 intdx + c_1`
∴ `x^2 y = x^5/5 + c_1`
∴ 5x2 y = x5 + c …[c = 5c1]
APPEARS IN
संबंधित प्रश्न
Verify that y2 = 4a (x + a) is a solution of the differential equations
\[y\left\{ 1 - \left( \frac{dy}{dx} \right)^2 \right\} = 2x\frac{dy}{dx}\]
Verify that y = log \[\left( x + \sqrt{x^2 + a^2} \right)^2\] satisfies the differential equation \[\left( a^2 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = 0\]
xy (y + 1) dy = (x2 + 1) dx
dy + (x + 1) (y + 1) dx = 0
Solve the following differential equation:
(xy2 + 2x) dx + (x2 y + 2y) dy = 0
Solve the differential equation \[\frac{dy}{dx} = \frac{2x\left( \log x + 1 \right)}{\sin y + y \cos y}\], given that y = 0, when x = 1.
Solve the following initial value problem:
\[\frac{dy}{dx} + y \cot x = 4x\text{ cosec }x, y\left( \frac{\pi}{2} \right) = 0\]
The rate of increase of bacteria in a culture is proportional to the number of bacteria present and it is found that the number doubles in 6 hours. Prove that the bacteria becomes 8 times at the end of 18 hours.
The integrating factor of the differential equation (x log x)
\[\frac{dy}{dx} + y = 2 \log x\], is given by
y2 dx + (x2 − xy + y2) dy = 0
In the following example, verify that the given function is a solution of the corresponding differential equation.
| Solution | D.E. |
| xy = log y + k | y' (1 - xy) = y2 |
Solve the following differential equation.
`dy/dx + y` = 3
Solve `("d"y)/("d"x) = (x + y + 1)/(x + y - 1)` when x = `2/3`, y = `1/3`
For the differential equation, find the particular solution (x – y2x) dx – (y + x2y) dy = 0 when x = 2, y = 0
Choose the correct alternative:
General solution of `y - x ("d"y)/("d"x)` = 0 is
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`
