Advertisements
Advertisements
प्रश्न
xy (y + 1) dy = (x2 + 1) dx
Advertisements
उत्तर
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
संबंधित प्रश्न
Show that Ax2 + By2 = 1 is a solution of the differential equation x \[\left\{ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right\} = y\frac{dy}{dx}\]
Show that y = ex (A cos x + B sin x) is the solution of the differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + 2y = 0\]
Verify that y = log \[\left( x + \sqrt{x^2 + a^2} \right)^2\] satisfies the differential equation \[\left( a^2 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = 0\]
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
Differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 2\] Function y = xex + ex
(sin x + cos x) dy + (cos x − sin x) dx = 0
C' (x) = 2 + 0.15 x ; C(0) = 100
x2 dy + y (x + y) dx = 0
Find the particular solution of the differential equation \[\frac{dy}{dx} = \frac{xy}{x^2 + y^2}\] given that y = 1 when x = 0.
Solve the following initial value problem:-
\[\frac{dy}{dx} + y \tan x = 2x + x^2 \tan x, y\left( 0 \right) = 1\]
In a culture, the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present?
Find the equation to the curve satisfying x (x + 1) \[\frac{dy}{dx} - y\] = x (x + 1) and passing through (1, 0).
The rate of increase of bacteria in a culture is proportional to the number of bacteria present and it is found that the number doubles in 6 hours. Prove that the bacteria becomes 8 times at the end of 18 hours.
Write the differential equation obtained by eliminating the arbitrary constant C in the equation x2 − y2 = C2.
The solution of the differential equation y1 y3 = y22 is
The differential equation satisfied by ax2 + by2 = 1 is
The differential equation \[x\frac{dy}{dx} - y = x^2\], has the general solution
Solve the following differential equation.
`dy/dx = x^2 y + y`
For each of the following differential equations find the particular solution.
(x − y2 x) dx − (y + x2 y) dy = 0, when x = 2, y = 0
Solve the following differential equation.
(x2 − y2 ) dx + 2xy dy = 0
Choose the correct alternative.
The differential equation of y = `k_1 + k_2/x` is
Solve the differential equation sec2y tan x dy + sec2x tan y dx = 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
Solve the differential equation `"dy"/"dx" + 2xy` = y
Solve: ydx – xdy = x2ydx.
