Advertisements
Advertisements
प्रश्न
Find the particular solution of the differential equation `e^xsqrt(1-y^2)dx+y/xdy=0` , given that y=1 when x=0
Advertisements
उत्तर
We have:
`e^xsqrt(1−y2)dx+y/x dy=0 `
`e^xsqrt(1−y2)dx=-y/x dy..........(1)`
Separating the variables in equation (1), we get:
`xe^xdx=-y/sqrt(1-y^2)dy.........(2)`
Integrating both sides of equation (2), we have:
`int xe^xdx=-inty/sqrt(1-y^2)dy ............(3)`
`Now,intxe^xdx=xe^x-e^x+C_1=e^x(x-1)+C_1.......(4)`
`"Let " I=-inty/sqrt(1-y^2)dy`
putting `1-y^2=t` we get,
`-2ydy=dt`
`-ydy=dt/2`
`I=1/2intdt/sqrtt`
`=1/2xx2t^(1/2)+C_2`
`=t^(1/2)+C_2`
`=(1-y^2)^(1/2)+C2.......(5)`
Putting the values in equation (3), we get
`e^x(x-1)+C_1=(1-y^2)^(1/2)+C_2`
`e^x(x-1)=(1-y^2)^(1/2)+C, "where " C=C_2-C_1.......(6)`
on putting y=1 and x=0 in equation (6) we get C=-1
The particular solution of the given differential equation is `e^x(x-1)=(1-y^2)-1`
APPEARS IN
संबंधित प्रश्न
Solve : 3ex tanydx + (1 +ex) sec2 ydy = 0
Also, find the particular solution when x = 0 and y = π.
Find the particular solution of differential equation:
`dy/dx=-(x+ycosx)/(1+sinx) " given that " y= 1 " when "x = 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:
x + y = tan–1y : y2 y′ + y2 + 1 = 0
Find the general solution of the differential equation `dy/dx + sqrt((1-y^2)/(1-x^2)) = 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.
Solve the differential equation `[e^(-2sqrtx)/sqrtx - y/sqrtx] dx/dy = 1 (x != 0).`
The solution of the differential equation \[\frac{dy}{dx} = 1 + x + y^2 + x y^2 , y\left( 0 \right) = 0\] is
The general solution of the differential equation \[\frac{y dx - x dy}{y} = 0\], is
Find the general solution of the differential equation \[x \cos \left( \frac{y}{x} \right)\frac{dy}{dx} = y \cos\left( \frac{y}{x} \right) + x .\]
\[\frac{dy}{dx} - y \cot x = cosec\ x\]
(x2 + 1) dy + (2y − 1) dx = 0
`y sec^2 x + (y + 7) tan x(dy)/(dx)=0`
\[\frac{dy}{dx} + 2y = \sin 3x\]
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:-
\[\frac{dy}{dx} + \left( \sec x \right) y = \tan x\]
Solve the following differential equation:-
\[\left( x + 3 y^2 \right)\frac{dy}{dx} = y\]
Find a particular solution of the following differential equation:- (x + y) dy + (x − y) dx = 0; y = 1 when x = 1
Find the equation of a curve passing through the point (0, 1). If the slope of the tangent to the curve at any point (x, y) is equal to the sum of the x-coordinate and the product of the x-coordinate and y-coordinate of that point.
Solve: `y + "d"/("d"x) (xy) = x(sinx + logx)`
tan–1x + tan–1y = c is the general solution of the differential equation ______.
Integrating factor of the differential equation `("d"y)/("d"x) + y tanx - secx` = 0 is ______.
y = aemx+ be–mx satisfies which of the following differential equation?
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 ______.
General solution of the differential equation of the type `("d"x)/("d"x) + "P"_1x = "Q"_1` is given by ______.
Find the general solution of the differential equation `x (dy)/(dx) = y(logy - logx + 1)`.
If the solution curve of the differential equation `(dy)/(dx) = (x + y - 2)/(x - y)` passes through the point (2, 1) and (k + 1, 2), k > 0, then ______.
