Advertisements
Advertisements
प्रश्न
Solve: `("d"y)/("d"x) = cos(x + y) + sin(x + y)`. [Hint: Substitute x + y = z]
Advertisements
उत्तर
Given that: `("d"y)/("d"x) = cos(x + y) + sin(x + y)`
Put x + y = v, on differentiating w.r.t. x, we get,
`1 + ("d"y)/("d"x) = "dv"/"dx"`
∴ `("d"y)/("d"x) = "dv"/"dx" - 1`
∴ `"dv"/"dx" - 1` = cos v + sin v
⇒ `"dv"/"dx"` = cos v + sin v + 1
⇒ `"dv"/(cos"v" + sin"v" + 1)` = dx
Integrating both sides, we have
`int "dv"/(cos"v" + sin"v" + 1) = int 1 . "d"x`
⇒ `int "dv"/(((1 - tan^2 "v"/2)/(1 + tan^2 "v"/2) + (2tan "v"/2)/(1 + tan^2 "v"/2) + 1)) = int 1. "d"x`
⇒ `int ((1 + tan^2 "v"/2))/(1 - tan^2 "v"/2 + 2 tan "v"/2 + 1 + tan^2 "v"/2) "dv" = int 1."d"x`
⇒ `int (sec^2 "v"/2)/(2 + 2 tan "v"/2) "dv" = int 1."d"x`
Put `2 + 2 tan "v"/2` = t
`2 * 1/2 sec^2 "v"/2 "dv"` = dt
⇒ `sec^2 "v"/2 "dv"` = dt
⇒ `int "dt"/"t" = int 1."d"x`
⇒ `log|"t"|` = x + c
⇒ `log|2 + 2 tan "v"/2|` = x + c
⇒ `log|2 + 2tan((x + y)/2)| ` = x + c
⇒ `log2 [1 + tan((x + y)/2)]` = x + c
⇒ `log2 + log[1 + tan ((x + y)/2)]` = x + c
⇒ `log[1 + tan((x + y)/2)]` = x + c – log 2
Hence, the required solution is `log[1 + tan((x + y)/2)]` = x + K ....[c – log 2 = K]
APPEARS IN
संबंधित प्रश्न
Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.
Show that y = AeBx is a solution of the differential equation
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 \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} = 0, y \left( 0 \right) = 2, y'\left( 0 \right) = 1\]
Function y = ex + 1
x2 dy + y (x + y) dx = 0
(y2 − 2xy) dx = (x2 − 2xy) dy
Solve the following initial value problem:-
\[x\frac{dy}{dx} - y = \left( x + 1 \right) e^{- x} , y\left( 1 \right) = 0\]
Solve the following initial value problem:-
\[\frac{dy}{dx} + y \tan x = 2x + x^2 \tan x, y\left( 0 \right) = 1\]
Solve the following initial value problem:
\[\frac{dy}{dx} + y \cot x = 4x\text{ cosec }x, y\left( \frac{\pi}{2} \right) = 0\]
The slope of the tangent at a point P (x, y) on a curve is \[\frac{- x}{y}\]. If the curve passes through the point (3, −4), find the equation of the curve.
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).
Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y\] sin x = 1, is
The solution of the differential equation \[\frac{dy}{dx} - \frac{y\left( x + 1 \right)}{x} = 0\] is given by
In the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:-
y = ex + 1 y'' − y' = 0
Find the coordinates of the centre, foci and equation of directrix of the hyperbola x2 – 3y2 – 4x = 8.
Solve the differential equation:
`"x"("dy")/("dx")+"y"=3"x"^2-2`
Find the particular solution of the differential equation `"dy"/"dx" = "xy"/("x"^2+"y"^2),`given that y = 1 when x = 0
Determine the order and degree of the following differential equations.
| Solution | D.E |
| y = aex + be−x | `(d^2y)/dx^2= 1` |
Solve the following differential equation.
`y^3 - dy/dx = x dy/dx`
Solve the following differential equation.
(x2 − y2 ) dx + 2xy dy = 0
Solve the following differential equation.
`xy dy/dx = x^2 + 2y^2`
Solve the following differential equation.
`x^2 dy/dx = x^2 +xy - y^2`
State whether the following is True or False:
The degree of a differential equation is the power of the highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any.
x2y dx – (x3 + y3) dy = 0
`dy/dx = log x`
Solve: `("d"y)/("d"x) + 2/xy` = x2
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`
Integrating factor of the differential equation `x "dy"/"dx" - y` = sinx is ______.
There are n students in a school. If r % among the students are 12 years or younger, which of the following expressions represents the number of students who are older than 12?
A man is moving away from a tower 41.6 m high at a rate of 2 m/s. If the eye level of the man is 1.6 m above the ground, then the rate at which the angle of elevation of the top of the tower changes, when he is at a distance of 30 m from the foot of the tower, is
