Advertisements
Advertisements
Question
`xy dy/dx = x^2 + 2y^2`
Advertisements
Solution
`xy dy/dx = x^2 + 2y^2`
∴ `dy/dx = x^2 + (2y^2)/ (xy)` …(i)
Put y = tx ...(ii)
Differentiating w.r.t. x, we get
`dy/dx = t + x dt/dx` ...(iii)
Substituting (ii) and (iii) in (i), we get
`t + x dt/dx = (x^2 + 2t^2 x^2)/ (x(tx))`
∴ `t + x dt/dx = (x^2 (1+2t^2))/(x^2t)`
∴ `x dt/dx (1+2t^2)/t - t = (1+ t^2)/t`
∴ `t/(1+t^2) dt = 1/x dx`
Integrating on both sides, we get
`1/2 int (2t)/(1+t^2) dt = int dx/x`
∴ `1/2 log|1 + t^2| = log |x| + log |c|`
∴ log |1 + t2 | = 2 log |x| + 2log |c|
= log |x2 | + log |c2|
∴ log |1 + t2 | = log |c2 x2|
∴ 1 + t2 = c2x2
∴ `1 + y^2/x^2 = c^2x^2`
∴ x2 + y2 = c2 x4
APPEARS IN
RELATED QUESTIONS
Verify that y = cx + 2c2 is a solution of the differential equation
xy dy = (y − 1) (x + 1) dx
dy + (x + 1) (y + 1) dx = 0
Solve the following differential equation:
\[\text{ cosec }x \log y \frac{dy}{dx} + x^2 y^2 = 0\]
Solve the following initial value problem:
\[x\frac{dy}{dx} + y = x \cos x + \sin x, y\left( \frac{\pi}{2} \right) = 1\]
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 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).
The x-intercept of the tangent line to a curve is equal to the ordinate of the point of contact. Find the particular curve through the point (1, 1).
Write the differential equation obtained eliminating the arbitrary constant C in the equation xy = C2.
If sin x is an integrating factor of the differential equation \[\frac{dy}{dx} + Py = Q\], then write the value of P.
The integrating factor of the differential equation \[\left( 1 - y^2 \right)\frac{dx}{dy} + yx = ay\left( - 1 < y < 1 \right)\] is ______.
If xmyn = (x + y)m+n, prove that \[\frac{dy}{dx} = \frac{y}{x} .\]
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.
Solve the following differential equation.
`y^3 - dy/dx = x dy/dx`
Solve the following differential equation.
`dy/dx + y` = 3
Solve:
(x + y) dy = a2 dx
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"` = yex and x = 0, y = e. Find the value of y when x = 1.
If `y = log_2 log_2(x)` then `(dy)/(dx)` =
`d/(dx)(tan^-1 (sqrt(1 + x^2) - 1)/x)` is equal to:
The differential equation (1 + y2)x dx – (1 + x2)y dy = 0 represents a family of:
