Advertisements
Advertisements
प्रश्न
Solve the following differential equation y2dx + (xy + x2) dy = 0
Advertisements
उत्तर
y2dx + (xy + x2) dy = 0
∴ `y^2 + (xy + x^2) ("d"y)/("d"x)` = 0
∴ `(xy + x^2) ("d"y)/("d"x)` = − y2
∴ `("d"y)/("d"x) = (-y^2)/(xy + x^2)` .....(i)
Put y = tx ......(ii)
Differentiating w.r.t. x, we get
`("d"y)/("d"x) = "t" + x "dt"/("d"x)` ......(iii)
Substituting (ii) and (iii) in (i), we get
`"t" + x "dt"/("d"x) = (-"t"^2x^2)/(x("t"x) + x^2)`
∴ `"t" + x "dt"/("d"x) = (-"t"^2x^2)/(x^2("t" + 1))`
∴ `"t" + x "dt"/("d"x) = (-"t"^2)/(1 + "t")`
∴ `x "dt"/("d"x) = (-"t"^2)/(1 + "t") - "t"`
= `(-"t"^2 - "t" - "t"^2)/(1 + "t")`
= `(-2"t"^2 - "t")/(1 + "t")`
∴ `(1 + "t")/("t"^2 + "t") "dt" = - ("d"x)/x`
Integrating on both sides, we get
`int (1 + "t")/(2"t"^2 + "t") "dt" = - int ("d"x)/x`
∴ `int ((2"t" + 1) - "t")/("t"(21"t" + 1)) "dt" = -int ("d"x)/x`
∴ `int (1/"t" - 1/(2"t" + 1)) "dt" = -int ("d"x)/x`
∴ `int 1/"t" "dt" - 1/2 int 2/(2"t" + 1) "dt" = -int ("d"x)/x`
∴ `log |"t"| - 1/2 log|2"t" + 1|` = − log|x| + log |c|
∴ `log |y/x| - 1/2 log|2(y/x) + 1|` = − log|x| + log |c|
∴ `log |y| - log |x| - 1/2 log|(2y + x)/x|` = − log|x| + log |c|
∴ `1/2 log|y^2| - 1/2 log|(2y + x)/x|` = log |c|
∴ `1/2 log|y^2/((2y + x)/x)|` = log |c|
∴ `1/2 log|(xy^2)/(2y + x)|` = log |c|
संबंधित प्रश्न
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 Ax2 + By2 = 1 is a solution of the differential equation x \[\left\{ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right\} = y\frac{dy}{dx}\]
Hence, the given function is the solution to the given differential equation. \[\frac{c - x}{1 + cx}\] is a solution of the differential equation \[(1+x^2)\frac{dy}{dx}+(1+y^2)=0\].
Verify that \[y = ce^{tan^{- 1}} x\] is a solution of the differential equation \[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + \left( 2x - 1 \right)\frac{dy}{dx} = 0\]
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x\frac{dy}{dx} + y = y^2\]
|
\[y = \frac{a}{x + a}\]
|
Differential equation \[x\frac{dy}{dx} = 1, y\left( 1 \right) = 0\]
Function y = log x
C' (x) = 2 + 0.15 x ; C(0) = 100
(ey + 1) cos x dx + ey sin x dy = 0
x cos2 y dx = y cos2 x dy
(1 + x) (1 + y2) dx + (1 + y) (1 + x2) dy = 0
tan y \[\frac{dy}{dx}\] = sin (x + y) + sin (x − y)
Solve the following differential equation:
\[y\left( 1 - x^2 \right)\frac{dy}{dx} = x\left( 1 + y^2 \right)\]
Solve the differential equation \[x\frac{dy}{dx} + \cot y = 0\] given that \[y = \frac{\pi}{4}\], when \[x=\sqrt{2}\]
Solve the differential equation \[\left( 1 + x^2 \right)\frac{dy}{dx} + \left( 1 + y^2 \right) = 0\], given that y = 1, when x = 0.
Find the particular solution of edy/dx = x + 1, given that y = 3, when x = 0.
Find the particular solution of the differential equation \[\frac{dy}{dx} = - 4x y^2\] given that y = 1, when x = 0.
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:-
\[dy = \cos x\left( 2 - y\text{ cosec }x \right)dx\]
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?
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?
If the interest is compounded continuously at 6% per annum, how much worth Rs 1000 will be after 10 years? How long will it take to double Rs 1000?
Find the curve for which the intercept cut-off by a tangent on x-axis is equal to four times the ordinate of the point of contact.
Find the equation of the curve which passes through the point (1, 2) and the distance between the foot of the ordinate of the point of contact and the point of intersection of the tangent with x-axis is twice the abscissa of the point of contact.
Show that all curves for which the slope at any point (x, y) on it is \[\frac{x^2 + y^2}{2xy}\] are rectangular hyperbola.
Find the equation of the curve passing through the point (0, 1) if the slope of the tangent to the curve at each of its point is equal to the sum of the abscissa and the product of the abscissa and the ordinate of the point.
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).
Write the differential equation obtained eliminating the arbitrary constant C in the equation xy = C2.
The solution of the differential equation y1 y3 = y22 is
Which of the following is the integrating factor of (x log x) \[\frac{dy}{dx} + y\] = 2 log x?
The integrating factor of the differential equation \[\left( 1 - y^2 \right)\frac{dx}{dy} + yx = ay\left( - 1 < y < 1 \right)\] is ______.
Verify that the function y = e−3x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + \frac{dy}{dx} - 6y = 0.\]
The price of six different commodities for years 2009 and year 2011 are as follows:
| Commodities | A | B | C | D | E | F |
|
Price in 2009 (₹) |
35 | 80 | 25 | 30 | 80 | x |
| Price in 2011 (₹) | 50 | y | 45 | 70 | 120 | 105 |
The Index number for the year 2011 taking 2009 as the base year for the above data was calculated to be 125. Find the values of x andy if the total price in 2009 is ₹ 360.
Find the particular solution of the differential equation `"dy"/"dx" = "xy"/("x"^2+"y"^2),`given that y = 1 when x = 0
In the following example, verify that the given function is a solution of the corresponding differential equation.
| Solution | D.E. |
| xy = log y + k | y' (1 - xy) = y2 |
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` |
Find the differential equation whose general solution is
x3 + y3 = 35ax.
Form the differential equation from the relation x2 + 4y2 = 4b2
For the following differential equation find the particular solution.
`dy/ dx = (4x + y + 1),
when y = 1, x = 0
Solve the following differential equation.
y2 dx + (xy + x2 ) dy = 0
Solve the following differential equation.
`dy /dx +(x-2 y)/ (2x- y)= 0`
Solve the following differential equation.
`(x + y) dy/dx = 1`
Choose the correct alternative.
The solution of `x dy/dx = y` log y is
A solution of a differential equation which can be obtained from the general solution by giving particular values to the arbitrary constants is called ___________ solution.
y2 dx + (xy + x2)dy = 0
Solve the differential equation `("d"y)/("d"x) + y` = e−x
For the differential equation, find the particular solution
`("d"y)/("d"x)` = (4x +y + 1), when y = 1, x = 0
Solve the following differential equation
`x^2 ("d"y)/("d"x)` = x2 + xy − y2
Choose the correct alternative:
Solution of the equation `x("d"y)/("d"x)` = y log y is
The function y = cx is the solution of differential equation `("d"y)/("d"x) = y/x`
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`
Given that `"dy"/"dx" = "e"^-2x` and y = 0 when x = 5. Find the value of x when y = 3.
Solve the differential equation `"dy"/"dx"` = 1 + x + y2 + xy2, when y = 0, x = 0.
lf the straight lines `ax + by + p` = 0 and `x cos alpha + y sin alpha = p` are inclined at an angle π/4 and concurrent with the straight line `x sin alpha - y cos alpha` = 0, then the value of `a^2 + b^2` is
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
