Advertisements
Advertisements
Question
Find the particular solution of the differential equation `(1+x^2)dy/dx=(e^(mtan^-1 x)-y)` , give that y=1 when x=0.
Solution
`(1+x^2)dy/dx=(e^(mtan^-1 x)-y)`
`=>dy/dx=(e^(mtav^-1 x))/(1+x^2)-y/(1+x^2)`
`=>dy/dx+y/(1+x^2)=e^(m tan^-1 x)/(1+x^2)`
`P=1/(1+x^2), Q=e^(m tan^-1 x)/(1+x^2)`
`I.F=e^(intPdx)`
`=e^(int(1/(1+x^2))dx)`
`=e^(tan^-1 x)`
Thus the solution is
`ye^(intPdx)=intQe^(intPdx)dx`
`=>yxxe^(tan^-1 x)=inte^(m tan^-1 x)/(1+x^2) .e^(tan^-1 x)dx`
`=>yxxe^(tan^-1 x)=inte^((m+1) tan^-1 x)/(1+x^2)dx ...............(i)`
`inte^((m+1) tan^-1 x)/(1+x^2)dx.............(ii)`
`Let (m+1)tan^-1 x = z`
`(m+1)/(1+x^2)dx=dz`
`dx/(1+x^2)=(dz)/(m+1)`
Substituting in (ii),
`1/(m+1)inte^z dz`
`=e^z/(m+1)`
`=e^((m+1)tan^-1 x)`
Substituting in (i),
`=>y xx e^(tan^-1 x)=e^((m+1)tan^-1 x)/(m+1)+C..........(iii)`
Putting y=1 and x=1, in the above equation,
`=>yxxe^(tan^-1 1)=e^((m+1)tan^-1 x)/(m+1)+C`
`=>1 xx e^(pi/4) = e^((m+1)tan^-1 pi/4)/(m+1)+C`
`=>C= e^((m+1)tan^-1 pi/4)/(m+1)- e^(pi/4)`
Particular solution of the D.E. is `yxxe^(tan^-1x)=e^((m+1)tan^-1 x)/(m+1)+e^((m+1)tan^-1 pi/4)/(m+1)- e^(pi/4)`
APPEARS IN
RELATED QUESTIONS
If `y=sqrt(sinx+sqrt(sinx+sqrt(sinx+..... oo))),` then show that `dy/dx=cosx/(2y-1)`
Find the particular solution of the differential equation `dy/dx=(xy)/(x^2+y^2)` given that y = 1, when x = 0.
Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
y = ex + 1 : y″ – y′ = 0
Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
y = x2 + 2x + C : y′ – 2x – 2 = 0
Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
y = Ax : xy′ = y (x ≠ 0)
Show that the general solution of the differential equation `dy/dx + (y^2 + y +1)/(x^2 + x + 1) = 0` is given by (x + y + 1) = A (1 - x - y - 2xy), where A is parameter.
The solution of the differential equation \[\frac{dy}{dx} = 1 + x + y^2 + x y^2 , y\left( 0 \right) = 0\] is
The solution of the differential equation \[2x\frac{dy}{dx} - y = 3\] represents
If m and n are the order and degree of the differential equation \[\left( y_2 \right)^5 + \frac{4 \left( y_2 \right)^3}{y_3} + y_3 = x^2 - 1\], then
Which of the following differential equations has y = x as one of its particular solution?
The general solution of the differential equation ex dy + (y ex + 2x) dx = 0 is
\[\frac{dy}{dx} + \frac{y}{x} = \frac{y^2}{x^2}\]
(x + y − 1) dy = (x + y) dx
Solve the following differential equation:- `y dx + x log (y)/(x)dy-2x dy=0`
Solve the following differential equation:-
\[\frac{dy}{dx} + 2y = \sin x\]
Solve the following differential equation:-
\[\frac{dy}{dx} + \frac{y}{x} = x^2\]
Solve the following differential equation:-
\[\left( x + y \right)\frac{dy}{dx} = 1\]
Solve the following differential equation:-
y dx + (x − y2) dy = 0
Find the equation of a curve passing through the point (−2, 3), given that the slope of the tangent to the curve at any point (x, y) is `(2x)/y^2.`
The solution of the differential equation `x "dt"/"dx" + 2y` = x2 is ______.
Solve the differential equation (1 + y2) tan–1xdx + 2y(1 + x2)dy = 0.
Solution of differential equation xdy – ydx = 0 represents : ______.
tan–1x + tan–1y = c is the general solution of the differential equation ______.
The general solution of ex cosy dx – ex siny dy = 0 is ______.
The solution of `x ("d"y)/("d"x) + y` = ex is ______.
The general solution of the differential equation `("d"y)/("d"x) = "e"^(x^2/2) + xy` is ______.
The solution of the differential equation `("d"y)/("d"x) = "e"^(x - y) + x^2 "e"^-y` is ______.
The solution of the differential equation ydx + (x + xy)dy = 0 is ______.
Solve the differential equation:
`(xdy - ydx) ysin(y/x) = (ydx + xdy) xcos(y/x)`.
Find the particular solution satisfying the condition that y = π when x = 1.
