Advertisements
Advertisements
प्रश्न
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
Advertisements
उत्तर
`("d"y)/("d"x)` = cos(x + y) ......(1)
Put x + y = v
∴ `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
∴ `1/(1 + cos "v")` 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 sec^2 ("v"/2) "dv" = int "d"x`
∴ `1/2* (tan("v"/2))/(1/2)` = x + c
∴ `tan ((x + y)/2)` = x + c
APPEARS IN
संबंधित प्रश्न
Verify that y = − x − 1 is a solution of the differential equation (y − x) dy − (y2 − x2) dx = 0.
Verify that y2 = 4a (x + a) is a solution of the differential equations
\[y\left\{ 1 - \left( \frac{dy}{dx} \right)^2 \right\} = 2x\frac{dy}{dx}\]
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x\frac{dy}{dx} = y\]
|
y = ax |
Differential equation \[\frac{dy}{dx} = y, y\left( 0 \right) = 1\]
Function y = ex
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 \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} = 0, y \left( 0 \right) = 2, y'\left( 0 \right) = 1\]
Function y = ex + 1
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
tan y dx + sec2 y tan x dy = 0
Solve the following differential equation:
\[\text{ cosec }x \log y \frac{dy}{dx} + x^2 y^2 = 0\]
The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after `t` seconds.
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:-
\[x\frac{dy}{dx} - y = \log x, y\left( 1 \right) = 0\]
The surface area of a balloon being inflated, changes at a rate proportional to time t. If initially its radius is 1 unit and after 3 seconds it is 2 units, find the radius after time t.
A bank pays interest by continuous compounding, that is, by treating the interest rate as the instantaneous rate of change of principal. Suppose in an account interest accrues at 8% per year, compounded continuously. Calculate the percentage increase in such an account over one year.
The differential equation of the ellipse \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = C\] is
The differential equation \[x\frac{dy}{dx} - y = x^2\], has the general solution
Show that y = ae2x + be−x is a solution of the differential equation \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} - 2y = 0\]
Form the differential equation representing the family of curves y = a sin (x + b), where a, b are arbitrary constant.
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` |
Solve the following differential equation.
x2y dx − (x3 + y3) dy = 0
Solve:
(x + y) dy = a2 dx
Solve the differential equation xdx + 2ydy = 0
Choose the correct alternative:
Solution of the equation `x("d"y)/("d"x)` = y log y is
