Advertisements
Advertisements
प्रश्न
उत्तर
We have,
\[\left( x^2 + 1 \right)\frac{dy}{dx} = 1\]
\[ \Rightarrow \frac{dy}{dx} = \frac{1}{x^2 + 1}\]
\[ \Rightarrow dy = \left( \frac{1}{x^2 + 1} \right)dx\]
Integrating both sides, we get
\[ \Rightarrow \int dy = \int\left( \frac{1}{x^2 + 1} \right)dx\]
\[ \Rightarrow y = \tan^{- 1} x + C\]
\[\text{ So,} y = \tan^{- 1} x + \text{ C is defined for all }x \in R . \]
\[\text{ Hence, }y = \tan^{- 1} x + \text{ C, where }x \in R, \text{ is the solution to the given differential equation }.\]
APPEARS IN
संबंधित प्रश्न
Show that y = ax3 + bx2 + c is a solution of the differential equation \[\frac{d^3 y}{d x^3} = 6a\].
Hence, the given function is the solution to the given differential equation. \[\frac{c - x}{1 + cx}\] is a solution of the differential equation \[(1+x^2)\frac{dy}{dx}+(1+y^2)=0\].
dy + (x + 1) (y + 1) dx = 0
Solve the following differential equation:
\[\text{ cosec }x \log y \frac{dy}{dx} + x^2 y^2 = 0\]
If y(x) is a solution of the different equation \[\left( \frac{2 + \sin x}{1 + y} \right)\frac{dy}{dx} = - \cos x\] and y(0) = 1, then find the value of y(π/2).
Solve the following initial value problem:-
\[x\frac{dy}{dx} - y = \left( x + 1 \right) e^{- x} , y\left( 1 \right) = 0\]
Solve the following initial value problem:-
\[dy = \cos x\left( 2 - y\text{ cosec }x \right)dx\]
Solve the following initial value problem:-
\[\tan x\left( \frac{dy}{dx} \right) = 2x\tan x + x^2 - y; \tan x \neq 0\] given that y = 0 when \[x = \frac{\pi}{2}\]
Find the curve for which the intercept cut-off by a tangent on x-axis is equal to four times the ordinate of the point of contact.
Show that the equation of the curve whose slope at any point is equal to y + 2x and which passes through the origin is y + 2 (x + 1) = 2e2x.
The integrating factor of the differential equation (x log x)
\[\frac{dy}{dx} + y = 2 \log x\], is given by
Solve the following differential equation : \[y^2 dx + \left( x^2 - xy + y^2 \right)dy = 0\] .
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.
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.
`(dθ)/dt = − k (θ − θ_0)`
Solve the following differential equation.
`dy /dx +(x-2 y)/ (2x- y)= 0`
Solve the differential equation xdx + 2ydy = 0
For the differential equation, find the particular solution
`("d"y)/("d"x)` = (4x +y + 1), when y = 1, x = 0
A solution of differential equation which can be obtained from the general solution by giving particular values to the arbitrary constant is called ______ solution
Verify y = log x + c is the solution of differential equation `x ("d"^2y)/("d"x^2) + ("d"y)/("d"x)` = 0
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`
Integrating factor of the differential equation `x "dy"/"dx" - y` = sinx is ______.
Given that `"dy"/"dx" = "e"^-2x` and y = 0 when x = 5. Find the value of x when y = 3.
