Advertisements
Advertisements
Question
For the following differential equation, find the general solution:- \[\frac{dy}{dx} = \left( 1 + x^2 \right)\left( 1 + y^2 \right)\]
Advertisements
Solution
We have,
\[\frac{dy}{dx} = \left( 1 + x^2 \right)\left( 1 + y^2 \right)\]
\[ \Rightarrow \frac{1}{\left( 1 + y^2 \right)}dy = \left( 1 + x^2 \right)dx\]
Integrating both sides, we get
\[\int\frac{1}{\left( 1 + y^2 \right)}dy = \int\left( 1 + x^2 \right)dx\]
\[ \Rightarrow \tan^{- 1} y = x + \frac{x^3}{3} + C\]
APPEARS IN
RELATED QUESTIONS
Solve the differential equation cos(x +y) dy = dx hence find the particular solution for x = 0 and y = 0.
If x = Φ(t) differentiable function of ‘ t ' then prove that `int f(x) dx=intf[phi(t)]phi'(t)dt`
Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes.
Find the particular solution of differential equation:
`dy/dx=-(x+ycosx)/(1+sinx) " given that " y= 1 " when "x = 0`
If y = P eax + Q ebx, show that
`(d^y)/(dx^2)=(a+b)dy/dx+aby=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:
xy = log y + C : `y' = (y^2)/(1 - xy) (xy != 1)`
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
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)`
Solve the differential equation `cos^2 x dy/dx` + y = tan x
Solve the differential equation:
`e^(x/y)(1-x/y) + (1 + e^(x/y)) dx/dy = 0` when x = 0, y = 1
The solution of the differential equation \[x\frac{dy}{dx} = y + x \tan\frac{y}{x}\], is
The solution of the differential equation \[\frac{dy}{dx} + 1 = e^{x + y}\], is
The solution of x2 + y2 \[\frac{dy}{dx}\]= 4, is
The solution of the differential equation \[\left( 1 + x^2 \right)\frac{dy}{dx} + 1 + y^2 = 0\], is
The general solution of the differential equation \[\frac{dy}{dx} = e^{x + y}\], is
The general solution of the differential equation ex dy + (y ex + 2x) dx = 0 is
\[\frac{dy}{dx} - y \cot x = cosec\ x\]
(x2 + 1) dy + (2y − 1) dx = 0
\[y - x\frac{dy}{dx} = b\left( 1 + x^2 \frac{dy}{dx} \right)\]
`x cos x(dy)/(dx)+y(x sin x + cos x)=1`
Solve the following differential equation:- \[\left( x - y \right)\frac{dy}{dx} = x + 2y\]
Solve the following differential equation:-
\[\frac{dy}{dx} + \left( \sec x \right) y = \tan x\]
Solve the following differential equation:-
\[x \log x\frac{dy}{dx} + y = \frac{2}{x}\log x\]
Solve the following differential equation:-
\[\left( x + 3 y^2 \right)\frac{dy}{dx} = y\]
Find the equation of a curve passing through the point (0, 0) and whose differential equation is \[\frac{dy}{dx} = e^x \sin x.\]
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 the differential equation: ` ("x" + 1) (d"y")/(d"x") = 2e^-"y" - 1; y(0) = 0.`
The solution of the differential equation `x "dt"/"dx" + 2y` = x2 is ______.
y = x is a particular solution of the differential equation `("d"^2y)/("d"x^2) - x^2 "dy"/"dx" + xy` = x.
If y(x) is a solution of `((2 + sinx)/(1 + y))"dy"/"dx"` = – cosx and y (0) = 1, then find the value of `y(pi/2)`.
The general solution of ex cosy dx – ex siny dy = 0 is ______.
The solution of `("d"y)/("d"x) + y = "e"^-x`, y(0) = 0 is ______.
The solution of the differential equation `("d"y)/("d"x) + (1 + y^2)/(1 + x^2)` is ______.
The differential equation for which y = acosx + bsinx is a solution, is ______.
Which of the following differential equations has `y = x` as one of its particular solution?
Find the general solution of the differential equation:
`(dy)/(dx) = (3e^(2x) + 3e^(4x))/(e^x + e^-x)`
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 ______.
