Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
We have,
\[\frac{dy}{dx} = x^5 + x^2 - \frac{2}{x}\]
\[ \Rightarrow dy = \left( x^5 + x^2 - \frac{2}{x} \right)dx\]
Integrating both sides, we get
\[ \Rightarrow \int dy = \int\left( x^5 + x^2 - \frac{2}{x} \right)dx\]
\[ \Rightarrow y = \frac{x^6}{6} + \frac{x^3}{3} - 2\log\left| x \right| + C\]
\[\text{Clearly, } y = \frac{x^6}{6} + \frac{x^3}{3} - 2\log\left| x \right| + \text{ C is defined for all }x \in R \text{ except }x = 0 . \]
\[\text{ Hence, }y = \frac{x^6}{6} + \frac{x^3}{3} - 2\log\left| x \right| + \text{ C, where } x \in R - \left\{ 0 \right\}, \text{ is the solution to the given differential equation.}\]
APPEARS IN
संबंधित प्रश्न
Verify that y = cx + 2c2 is a solution of the differential equation
xy (y + 1) dy = (x2 + 1) dx
Solve the following differential equation:
\[y e^\frac{x}{y} dx = \left( x e^\frac{x}{y} + y^2 \right)dy, y \neq 0\]
In a bank principal increases at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0.5 = 1.648).
2xy dx + (x2 + 2y2) dy = 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:-
\[y' + y = e^x , y\left( 0 \right) = \frac{1}{2}\]
In a simple circuit of resistance R, self inductance L and voltage E, the current `i` at any time `t` is given by L \[\frac{di}{dt}\]+ R i = E. If E is constant and initially no current passes through the circuit, prove that \[i = \frac{E}{R}\left\{ 1 - e^{- \left( R/L \right)t} \right\}.\]
Find the equation of the curve which passes through the point (2, 2) and satisfies the differential equation
\[y - x\frac{dy}{dx} = y^2 + \frac{dy}{dx}\]
Show that the equation of the curve whose slope at any point is equal to y + 2x and which passes through the origin is y + 2 (x + 1) = 2e2x.
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 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.
The x-intercept of the tangent line to a curve is equal to the ordinate of the point of contact. Find the particular curve through the point (1, 1).
Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y\] sin x = 1, is
The solution of the differential equation \[\frac{dy}{dx} = \frac{ax + g}{by + f}\] represents a circle when
Solve the following differential equation : \[\left( \sqrt{1 + x^2 + y^2 + x^2 y^2} \right) dx + xy \ dy = 0\].
Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
Solve the differential equation:
`"x"("dy")/("dx")+"y"=3"x"^2-2`
For the following differential equation find the particular solution.
`(x + 1) dy/dx − 1 = 2e^(−y)`,
when y = 0, x = 1
Solve the following differential equation.
xdx + 2y dx = 0
Solve the following differential equation.
dr + (2r)dθ= 8dθ
Choose the correct alternative.
The differential equation of y = `k_1 + k_2/x` is
Solve the differential equation:
`e^(dy/dx) = x`
x2y dx – (x3 + y3) dy = 0
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 (x2 – yx2)dy + (y2 + xy2)dx = 0
The solution of differential equation `x^2 ("d"^2y)/("d"x^2)` = 1 is ______
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
Find the particular solution of the following differential equation
`("d"y)/("d"x)` = e2y cos x, when x = `pi/6`, y = 0.
Solution: The given D.E. is `("d"y)/("d"x)` = e2y cos x
∴ `1/"e"^(2y) "d"y` = cos x dx
Integrating, we get
`int square "d"y` = cos x dx
∴ `("e"^(-2y))/(-2)` = sin x + c1
∴ e–2y = – 2sin x – 2c1
∴ `square` = c, where c = – 2c1
This is general solution.
When x = `pi/6`, y = 0, we have
`"e"^0 + 2sin pi/6` = c
∴ c = `square`
∴ particular solution is `square`
`d/(dx)(tan^-1 (sqrt(1 + x^2) - 1)/x)` is equal to:
