Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
`(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
संबंधित प्रश्न
If `y=sqrt(sinx+sqrt(sinx+sqrt(sinx+..... oo))),` then show that `dy/dx=cosx/(2y-1)`
The differential equation of the family of curves y=c1ex+c2e-x is......
(a)`(d^2y)/dx^2+y=0`
(b)`(d^2y)/dx^2-y=0`
(c)`(d^2y)/dx^2+1=0`
(d)`(d^2y)/dx^2-1=0`
Find the differential equation representing the curve y = cx + c2.
Find the particular solution of differential equation:
`dy/dx=-(x+ycosx)/(1+sinx) " given that " y= 1 " when "x = 0`
Solve the differential equation `dy/dx -y =e^x`
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)
Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
`y = sqrt(a^2 - x^2 ) x in (-a,a) : x + y dy/dx = 0(y != 0)`
The number of arbitrary constants in the particular solution of a differential equation of third order are ______.
Write the order of the differential equation associated with the primitive y = C1 + C2 ex + C3 e−2x + C4, where C1, C2, C3, C4 are arbitrary constants.
The solution of the differential equation \[\frac{dy}{dx} = 1 + x + y^2 + x y^2 , y\left( 0 \right) = 0\] is
Solution of the differential equation \[\frac{dy}{dx} + \frac{y}{x}=\sin x\] is
The solution of x2 + y2 \[\frac{dy}{dx}\]= 4, is
The general solution of the differential equation \[\frac{dy}{dx} = e^{x + y}\], is
\[\frac{dy}{dx} = \left( x + y \right)^2\]
\[\frac{dy}{dx} - y \cot x = cosec\ x\]
\[\cos^2 x\frac{dy}{dx} + y = \tan x\]
`(dy)/(dx)+ y tan x = x^n cos x, n ne− 1`
Find the particular solution of the differential equation \[\frac{dy}{dx} = - 4x y^2\] given that y = 1, when x = 0.
For the following differential equation, find the general solution:- `y log y dx − x dy = 0`
Solve the following differential equation:-
\[x \log x\frac{dy}{dx} + y = \frac{2}{x}\log x\]
Find the equation of a curve passing through the point (0, 0) and whose differential equation is \[\frac{dy}{dx} = e^x \sin x.\]
The solution of the differential equation `x "dt"/"dx" + 2y` = x2 is ______.
Solve: `y + "d"/("d"x) (xy) = x(sinx + logx)`
The number of solutions of `("d"y)/("d"x) = (y + 1)/(x - 1)` when y (1) = 2 is ______.
The integrating factor of the differential equation `("d"y)/("d"x) + y = (1 + y)/x` is ______.
The general solution of `("d"y)/("d"x) = 2x"e"^(x^2 - y)` is ______.
Number of arbitrary constants in the particular solution of a differential equation of order two is two.
The solution of the differential equation `("d"y)/("d"x) = (x + 2y)/x` is x + y = kx2.
