Advertisements
Advertisements
Question
xy (y + 1) dy = (x2 + 1) dx
Advertisements
Solution
We have,
\[xy\left( y + 1 \right)dy = \left( x^2 + 1 \right)dx\]
\[ \Rightarrow \left\{ y\left( y + 1 \right) \right\}dy = \frac{x^2 + 1}{x}dx\]
\[ \Rightarrow \left( y^2 + y \right)dy = \left( x + \frac{1}{x} \right)dx\]
Integrating both sides, we get
\[\int\left( y^2 + y \right)dy = \int\left( x + \frac{1}{x} \right)dx\]
\[ \Rightarrow \int y^2 dy + \int y dy = \int x dx + \int\frac{1}{x}dx\]
\[ \Rightarrow \frac{y^3}{3} + \frac{y^2}{2} = \frac{x^2}{2} + \log \left| x \right| + C\]
\[\text{ Hence,} \frac{y^3}{3} + \frac{y^2}{2} = \frac{x^2}{2} + \log \left| x \right| +\text{ C is the required solution }. \]
APPEARS IN
RELATED QUESTIONS
Differential equation \[\frac{d^2 y}{d x^2} + y = 0, y \left( 0 \right) = 0, y' \left( 0 \right) = 1\] Function y = sin x
Differential equation \[\frac{dy}{dx} + y = 2, y \left( 0 \right) = 3\] Function y = e−x + 2
Differential equation \[\frac{d^2 y}{d x^2} - y = 0, y \left( 0 \right) = 2, y' \left( 0 \right) = 0\] Function y = ex + e−x
tan y \[\frac{dy}{dx}\] = sin (x + y) + sin (x − y)
Solve the following differential equation:
\[\left( 1 + y^2 \right) \tan^{- 1} xdx + 2y\left( 1 + x^2 \right)dy = 0\]
Find the solution of the differential equation cos y dy + cos x sin y dx = 0 given that y = \[\frac{\pi}{2}\], when x = \[\frac{\pi}{2}\]
Find the particular solution of the differential equation \[\frac{dy}{dx} = - 4x y^2\] given that y = 1, when x = 0.
y ex/y dx = (xex/y + y) dy
Solve the following initial value problem:-
\[x\frac{dy}{dx} - y = \log x, y\left( 1 \right) = 0\]
Solve the following initial value problem:-
\[dy = \cos x\left( 2 - y\text{ cosec }x \right)dx\]
Find the curve for which the intercept cut-off by a tangent on x-axis is equal to four times the ordinate of the point of contact.
Write the differential equation obtained eliminating the arbitrary constant C in the equation xy = C2.
The solution of the differential equation y1 y3 = y22 is
The differential equation of the ellipse \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = C\] is
Find the coordinates of the centre, foci and equation of directrix of the hyperbola x2 – 3y2 – 4x = 8.
In the following example, verify that the given function is a solution of the corresponding differential equation.
| Solution | D.E. |
| y = xn | `x^2(d^2y)/dx^2 - n xx (xdy)/dx + ny =0` |
In each of the following examples, verify that the given function is a solution of the corresponding differential equation.
| Solution | D.E. |
| y = ex | `dy/ dx= y` |
The solution of `dy/dx + x^2/y^2 = 0` is ______
Choose the correct alternative.
The solution of `x dy/dx = y` log y is
A solution of a differential equation which can be obtained from the general solution by giving particular values to the arbitrary constants is called ___________ solution.
Select and write the correct alternative from the given option for the question
Differential equation of the function c + 4yx = 0 is
Solve the differential equation `("d"y)/("d"x) + y` = e−x
Solve the differential equation xdx + 2ydy = 0
For the differential equation, find the particular solution (x – y2x) dx – (y + x2y) dy = 0 when x = 2, y = 0
Solve the following differential equation `("d"y)/("d"x)` = cos(x + y)
Solution: `("d"y)/("d"x)` = cos(x + y) ......(1)
Put `square`
∴ `1 + ("d"y)/("d"x) = "dv"/("d"x)`
∴ `("d"y)/("d"x) = "dv"/("d"x) - 1`
∴ (1) becomes `"dv"/("d"x) - 1` = cos v
∴ `"dv"/("d"x)` = 1 + cos v
∴ `square` dv = dx
Integrating, we get
`int 1/(1 + cos "v") "d"v = int "d"x`
∴ `int 1/(2cos^2 ("v"/2)) "dv" = int "d"x`
∴ `1/2 int square "dv" = int "d"x`
∴ `1/2* (tan("v"/2))/(1/2)` = x + c
∴ `square` = x + c
In mathematics and science, what primary purpose do differential equations serve?
