Advertisements
Advertisements
Question
For the differential equation, find the particular solution
`("d"y)/("d"x)` = (4x +y + 1), when y = 1, x = 0
Advertisements
Solution
`("d"y)/("d"x)` = (4x + y + 1) ......(i)
Put 4x + y + 1 = t .....(ii)
Differentiating w.r.t. x, we get
`4 + ("d"y)/("d"x) = ("dt")/("d"x)`
∴ `("d"y)/("d"x) = "dt"/("d"x) - 4` ......(iii)
Substituting (ii) and (iii) in (i), we get
`"dt"/("d"x) - 4` = t
∴ `"dt"/("d"x)` = u + 4
∴ `"dt"/("t" + 4)` = dx
Integrating on both sides, we get
`int "dt"/("t" + 4) = int "d"x`
∴ log |t + 4| = x + c
∴ log |(4x + y + 1) + 4| = x + c
∴ log |4x + y + 5| = x + c ......(iv)
When y = 1, x = 0
∴ log |4(0) + 1 + 5| = x + c
∴ c = log |6|
∴ log |4x + y + 5| = x + log 6 ....[From (iv)]
∴ log |4x + y + 5| – log 6 = x
∴ `log|(4x + y + 5)/6|` = x, which is the required particular solution
RELATED QUESTIONS
Solve the equation for x: `sin^(-1) 5/x + sin^(-1) 12/x = π/2, x ≠ 0`
Verify that y = 4 sin 3x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + 9y = 0\]
Show that the function y = A cos 2x − B sin 2x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + 4y = 0\].
Verify that y = \[\frac{a}{x} + b\] is a solution of the differential equation
\[\frac{d^2 y}{d x^2} + \frac{2}{x}\left( \frac{dy}{dx} \right) = 0\]
Show that y = e−x + ax + b is solution of the differential equation\[e^x \frac{d^2 y}{d x^2} = 1\]
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x + y\frac{dy}{dx} = 0\]
|
\[y = \pm \sqrt{a^2 - x^2}\]
|
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
xy (y + 1) dy = (x2 + 1) dx
(1 − x2) dy + xy dx = xy2 dx
tan y \[\frac{dy}{dx}\] = sin (x + y) + sin (x − y)
(y2 + 1) dx − (x2 + 1) dy = 0
Solve the following differential equation:
\[\text{ cosec }x \log y \frac{dy}{dx} + x^2 y^2 = 0\]
Solve the following differential equation:
\[xy\frac{dy}{dx} = 1 + x + y + xy\]
Solve the following differential equation:
\[y\left( 1 - x^2 \right)\frac{dy}{dx} = x\left( 1 + y^2 \right)\]
Solve the following differential equation:
\[\left( 1 + y^2 \right) \tan^{- 1} xdx + 2y\left( 1 + x^2 \right)dy = 0\]
Solve the differential equation \[x\frac{dy}{dx} + \cot y = 0\] given that \[y = \frac{\pi}{4}\], when \[x=\sqrt{2}\]
Find the particular solution of the differential equation \[\frac{dy}{dx} = - 4x y^2\] given that y = 1, when x = 0.
In a culture the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present.
y ex/y dx = (xex/y + y) dy
(y2 − 2xy) dx = (x2 − 2xy) dy
3x2 dy = (3xy + y2) dx
Solve the following initial value problem:-
\[x\frac{dy}{dx} - y = \log x, y\left( 1 \right) = 0\]
Solve the following initial value problem:-
\[\frac{dy}{dx} - 3y \cot x = \sin 2x; y = 2\text{ when }x = \frac{\pi}{2}\]
The rate of growth of a population is proportional to the number present. If the population of a city doubled in the past 25 years, and the present population is 100000, when will the city have a population of 500000?
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.
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.
The normal to a given curve at each point (x, y) on the curve passes through the point (3, 0). If the curve contains the point (3, 4), find its equation.
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 slope of a curve at each of its points is equal to the square of the abscissa of the point. Find the particular curve through the point (−1, 1).
Find the equation of the curve that passes through the point (0, a) and is such that at any point (x, y) on it, the product of its slope and the ordinate is equal to the abscissa.
Write the differential equation obtained by eliminating the arbitrary constant C in the equation x2 − y2 = C2.
Find the solution of the differential equation
\[x\sqrt{1 + y^2}dx + y\sqrt{1 + x^2}dy = 0\]
Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y\] sin x = 1, is
Which of the following is the integrating factor of (x log x) \[\frac{dy}{dx} + y\] = 2 log x?
If xmyn = (x + y)m+n, prove that \[\frac{dy}{dx} = \frac{y}{x} .\]
Show that y = ae2x + be−x is a solution of the differential equation \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} - 2y = 0\]
y2 dx + (x2 − xy + y2) dy = 0
Verify that the function y = e−3x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + \frac{dy}{dx} - 6y = 0.\]
Find the equation of the plane passing through the point (1, -2, 1) and perpendicular to the line joining the points A(3, 2, 1) and B(1, 4, 2).
Determine the order and degree of the following differential equations.
| Solution | D.E |
| y = aex + be−x | `(d^2y)/dx^2= 1` |
Form the differential equation from the relation x2 + 4y2 = 4b2
Solve the following differential equation.
`dy/dx + 2xy = x`
`xy dy/dx = x^2 + 2y^2`
Solve the differential equation `("d"y)/("d"x) + y` = e−x
Solve: `("d"y)/("d"x) + 2/xy` = x2
Solve the following differential equation y2dx + (xy + x2) dy = 0
A solution of differential equation which can be obtained from the general solution by giving particular values to the arbitrary constant is called ______ solution
The function y = ex is solution ______ of differential equation
Solve the following differential equation
sec2 x tan y dx + sec2 y tan x dy = 0
Solution: sec2 x tan y dx + sec2 y tan x dy = 0
∴ `(sec^2x)/tanx "d"x + square` = 0
Integrating, we get
`square + int (sec^2y)/tany "d"y` = log c
Each of these integral is of the type
`int ("f'"(x))/("f"(x)) "d"x` = log |f(x)| + log c
∴ the general solution is
`square + log |tan y|` = log c
∴ log |tan x . tan y| = log c
`square`
This is the general solution.
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
Solve `x^2 "dy"/"dx" - xy = 1 + cos(y/x)`, x ≠ 0 and x = 1, y = `pi/2`
Solution of `x("d"y)/("d"x) = y + x tan y/x` is `sin(y/x)` = cx
If `y = log_2 log_2(x)` then `(dy)/(dx)` =
