Advertisements
Advertisements
प्रश्न
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`
Advertisements
उत्तर
The given D.E. is `("d"y)/("d"x)` = e2y cos x
∴ `1/"e"^(2y) "d"y` = cos x dx
Integrating, we get
`int bb("e"^(-2y)) "d"y` = cos x dx
∴ `("e"^(-2y))/(-2)` = sin x + c1
∴ e–2y = – 2sin x – 2c1
∴ e–2y + 2sin x = c, where c = – 2c1
This is general solution.
When x = `pi/6`, y = 0, we have
`"e"^0 + 2sin pi/6` = c
∴ `1 + 2(1/2)` = c
∴ c = 2
∴ particular solution is e–2y + 2sin x = 2
APPEARS IN
संबंधित प्रश्न
Verify that y2 = 4a (x + a) is a solution of the differential equations
\[y\left\{ 1 - \left( \frac{dy}{dx} \right)^2 \right\} = 2x\frac{dy}{dx}\]
Differential equation \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} = 0, y \left( 0 \right) = 2, y'\left( 0 \right) = 1\]
Function y = ex + 1
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
(1 – y2) (1 + log x) dx + 2xy dy = 0, given that y = 0 when x = 1.
Find the equation of the curve which passes through the point (3, −4) and has the slope \[\frac{2y}{x}\] at any point (x, y) on it.
Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y\] sin x = 1, is
The differential equation \[x\frac{dy}{dx} - y = x^2\], has the general solution
The differential equation
\[\frac{dy}{dx} + Py = Q y^n , n > 2\] can be reduced to linear form by substituting
Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y \sin x = 1\], is
Form the differential equation of the family of circles having centre on y-axis and radius 3 unit.
Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
Find the coordinates of the centre, foci and equation of directrix of the hyperbola x2 – 3y2 – 4x = 8.
Find the particular solution of the differential equation `"dy"/"dx" = "xy"/("x"^2+"y"^2),`given that y = 1 when x = 0
Find the differential equation whose general solution is
x3 + y3 = 35ax.
Solve the following differential equation.
y dx + (x - y2 ) dy = 0
Solve `("d"y)/("d"x) = (x + y + 1)/(x + y - 1)` when x = `2/3`, y = `1/3`
Solve the following differential equation y log y = `(log y - x) ("d"y)/("d"x)`
Choose the correct alternative:
Solution of the equation `x("d"y)/("d"x)` = y log y is
Choose the correct alternative:
Differential equation of the function c + 4yx = 0 is
Solve the following differential equation
`y log y ("d"x)/("d"y) + x` = log 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
Given that `"dy"/"dx" = "e"^-2x` and y = 0 when x = 5. Find the value of x when y = 3.
Why is the equation \[x\frac{dy}{dx} + y = 0\] classified as a differential equation?
In mathematics and science, what primary purpose do differential equations serve?
