Advertisements
Advertisements
प्रश्न
Find the particular solution of the differential equation
`tan x * (dy)/(dx) = 2x tan x + x^2 - y`; `(tan x != 0)` given that y = 0 when `x = pi/2`
Advertisements
उत्तर
The given differential equation is
`tan x * (dy)/(dx) = 2x tan x + x^2 - y`; `(tan x != 0)`
`=> (dy)/(dx) = 2x + x^2 cotx - y cotx`
`=> dy/dx + (cot x)y = 2x + x^2 cot x`
This is a linear differential equation.
Here, P = cot x, Q = 2x + x2 cot x
:. I.F. = `e^(int P dx) = e^(int cot s dx) = e^(log|sin x|) = sin x`
The general solution of this linear differential equation is given by
y(I.F.) = ∫Q(I.F.)dx + C
`=> y*sinx = int(2x + x^2 cotx) sinx dx + C`
`=>y*sinx = int 2xsin x dx + int x^2 cos x dx + C`
`y*sinx = int2x sin x dx + x^2 sinx - int 2 xsin x + C` (Applying integration by parts in the 2nd integral)
`=>y*sinx = x^2 sinx +C`......1
When y = 0, `x = pi/2` (Given)
`:. 0 xx sin pi/2 = pi^2/4 sin pi/4 + C`
`=> C = - pi^2/4`
Substituting the value of C in (1), we get
`ysinx = x^2 sin x - pi^2/4`
`=> (x^2- y) sin x = pi^2/4`
This is the particular solution of the given differential equation
APPEARS IN
संबंधित प्रश्न
Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
y = cos x + C : y′ + sin x = 0
If y = etan x+ (log x)tan x then find dy/dx
The general solution of the differential equation \[\frac{dy}{dx} + y \] cot x = cosec x, is
The solution of x2 + y2 \[\frac{dy}{dx}\]= 4, is
The solution of the differential equation x dx + y dy = x2 y dy − y2 x dx, is
The solution of the differential equation \[\frac{dy}{dx} - ky = 0, y\left( 0 \right) = 1\] approaches to zero when x → ∞, if
The number of arbitrary constants in the general solution of differential equation of fourth order is
The solution of the differential equation \[\frac{dy}{dx} = \frac{y}{x} + \frac{\phi\left( \frac{y}{x} \right)}{\phi'\left( \frac{y}{x} \right)}\] is
(x + y − 1) dy = (x + y) dx
`(2ax+x^2)(dy)/(dx)=a^2+2ax`
x2 dy + (x2 − xy + y2) dx = 0
\[\frac{dy}{dx} + 5y = \cos 4x\]
\[x\frac{dy}{dx} + x \cos^2 \left( \frac{y}{x} \right) = y\]
For the following differential equation, find the general solution:- `y log y dx − x dy = 0`
For the following differential equation, find the general solution:- \[\frac{dy}{dx} + y = 1\]
Solve the following differential equation:-
\[\frac{dy}{dx} + 3y = e^{- 2x}\]
Find the equation of a curve passing through the point (0, 0) and whose differential equation is \[\frac{dy}{dx} = e^x \sin x.\]
Solve the differential equation: `(d"y")/(d"x") - (2"x")/(1+"x"^2) "y" = "x"^2 + 2`
The general solution of the differential equation `"dy"/"dx" = "e"^(x - y)` is ______.
The general solution of the differential equation `"dy"/"dx" + y sec x` = tan x is y(secx – tanx) = secx – tanx + x + k.
x + y = tan–1y is a solution of the differential equation `y^2 "dy"/"dx" + y^2 + 1` = 0.
Find the general solution of `(x + 2y^3) "dy"/"dx"` = y
If y = e–x (Acosx + Bsinx), then y is a solution of ______.
Solution of differential equation xdy – ydx = 0 represents : ______.
Integrating factor of `(x"d"y)/("d"x) - y = x^4 - 3x` is ______.
The solution of `("d"y)/("d"x) + y = "e"^-x`, y(0) = 0 is ______.
Find the particular solution of the differential equation `x (dy)/(dx) - y = x^2.e^x`, given y(1) = 0.
