Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
We have,
\[\left( x + 2 \right)\frac{dy}{dx} = x^2 + 3x + 7\]
\[ \Rightarrow \frac{dy}{dx} = \frac{x^2 + 3x + 7}{x + 2}\]
\[ \Rightarrow dy = \left( \frac{x^2 + 3x + 7}{x + 2} \right)dx\]
Integrating both sides, we get
\[\int dy = \int\left( \frac{x^2 + 3x + 7}{x + 2} \right)dx\]
\[ \Rightarrow \int dy = \int\left( \frac{x^2 + 3x + 2 + 5}{x + 2} \right)dx\]
\[ \Rightarrow \int dy = \int\left[ \frac{\left( x + 2 \right)\left( x + 1 \right) + 5}{x + 2} \right]dx\]
\[ \Rightarrow \int dy = \int\left( x + 1 + \frac{5}{x + 2} \right)dx\]
\[ \Rightarrow y = \frac{x^2}{2} + x + 5 \log\left| x + 2 \right| + C\]
\[\text{ So, } y = \frac{x^2}{2} + x + 5 \log\left| x + 2 \right| +\text{C is defined for all } x \in R\text{ except }x = - 2 . \]
\[\text{Hence, }y = \frac{x^2}{2} + x + 5 \log\left| x + 2 \right| + \text{C, where }x \in R - \left\{ 2 \right\},\text{ is the solution to the given differential equation.}\]
APPEARS IN
संबंधित प्रश्न
Verify that y = \[\frac{a}{x} + b\] is a solution of the differential equation
\[\frac{d^2 y}{d x^2} + \frac{2}{x}\left( \frac{dy}{dx} \right) = 0\]
Verify that y = − x − 1 is a solution of the differential equation (y − x) dy − (y2 − x2) dx = 0.
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[y = \left( \frac{dy}{dx} \right)^2\]
|
\[y = \frac{1}{4} \left( x \pm a \right)^2\]
|
(sin x + cos x) dy + (cos x − sin x) dx = 0
C' (x) = 2 + 0.15 x ; C(0) = 100
(1 − x2) dy + xy dx = xy2 dx
(y2 + 1) dx − (x2 + 1) dy = 0
Solve the following differential equation:
(xy2 + 2x) dx + (x2 y + 2y) dy = 0
Solve the following differential equations:
\[\frac{dy}{dx} = \frac{y}{x}\left\{ \log y - \log x + 1 \right\}\]
\[\frac{dy}{dx} = \frac{y}{x} + \sin\left( \frac{y}{x} \right)\]
Solve the following initial value problem:
\[\frac{dy}{dx} + y \cot x = 4x\text{ cosec }x, y\left( \frac{\pi}{2} \right) = 0\]
Solve the following initial value problem:-
\[\frac{dy}{dx} + y\cot x = 2\cos x, y\left( \frac{\pi}{2} \right) = 0\]
Experiments show that radium disintegrates at a rate proportional to the amount of radium present at the moment. Its half-life is 1590 years. What percentage will disappear in one year?
Find the equation of the curve passing through the point \[\left( 1, \frac{\pi}{4} \right)\] and tangent at any point of which makes an angle tan−1 \[\left( \frac{y}{x} - \cos^2 \frac{y}{x} \right)\] with x-axis.
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.
Find the equation of the curve such that the portion of the x-axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2).
The solution of the differential equation \[\frac{dy}{dx} = \frac{ax + g}{by + f}\] represents a circle when
Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of x-axis.
Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
Find the differential equation whose general solution is
x3 + y3 = 35ax.
Solve the differential equation sec2y tan x dy + sec2x tan y dx = 0
Solve: `("d"y)/("d"x) + 2/xy` = x2
State whether the following statement is True or False:
The integrating factor of the differential equation `("d"y)/("d"x) - y` = x is e–x
The function y = cx is the solution of differential equation `("d"y)/("d"x) = y/x`
Solve the following differential equation `("d"y)/("d"x)` = x2y + y
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
`d/(dx)(tan^-1 (sqrt(1 + x^2) - 1)/x)` is equal to:
