Advertisements
Advertisements
प्रश्न
Solve the following differential equation.
`dy /dx +(x-2 y)/ (2x- y)= 0`
उत्तर
`dy /dx +(x-2 y)/ (2x- y)= 0` ....(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-2tx)/(2x-tx) = 0`
∴`x dt/dx +t + (1-2t)/2-t = 0`
∴`x dt/dx + (2t - t^2+1-2t)/2-t = 0`
∴`x dt/dx + (1-t^2)/(2-t )= 0`
∴ `x dt/dx = - (1-t^2)/(2-t )`
∴ = `(2-t)/(1-t^2)dt = dx/x`
∴ `(2-t)/(t^2-1)dt = dx/x`
Integrating on both sides, we get
`int (2-t)/(t^2-1) dt = int dx/x`
∴ `int (2-t)/((t+1)(t-1)) dt = int dx/x`
Let `2-t/((t+1)(t-1)) = A/(t+1)+ B/(t-1)`
∴ 2 - t = A(t -1) + B(t + 1)
Putting t = 1, we get
∴ 2 -1 = A(1 -1) + B(1 + 1)
∴ B = `1 /2`
Putting t = -1, we get
2 -(-1) = A(-1 -1) + B(-1 + 1)
∴ A = `(-3)/2`
∴ `int(-3/2)/(t+1)dt +int(1/2)/(t-1) dt = intdx/x`
∴`(-3)/2 int 1/(t+1)dt + 1/2int 1/(t-1) dt = int dx/x`
∴`(-3)/2 log|t+1| + 1/2 log |t-1| = log |x| + log |c_1|`
∴ `-3 log |(y+x)/x| + log|(y-x)/x| = 2log |x| + 2 log |c_1|`
∴ -3 log |y+x| + 3 log |x| + log | y -x| - log |x|
= 2 log |x| + 2 log |c1|
∴ log |y - x| = 3 log |y+x|+ 2 log |c1|
∴ log |y- x |= log |( y+ x )3|+ log | c12|
∴ log | y - x| = log | c12 ( x+y)3|
∴ (y - x) = c(x + y) 3 … |c12 c|
Notes
Answer given in the textbook is `log |(x+y)/(x-y)| - 1/2 log | x^2 - y^2| + 2 log x = log c.`
However, as per our calculation it is ‘(y -x) = c(x+y)3.
APPEARS IN
संबंधित प्रश्न
Show that the function y = A cos x + B sin x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + y = 0\]
Show that the differential equation of which \[y = 2\left( x^2 - 1 \right) + c e^{- x^2}\] is a solution is \[\frac{dy}{dx} + 2xy = 4 x^3\]
Differential equation \[\frac{d^2 y}{d x^2} - 3\frac{dy}{dx} + 2y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 3\] Function y = ex + e2x
x cos2 y dx = y cos2 x dy
y (1 + ex) dy = (y + 1) ex dx
Find the equation of the curve passing through the point \[\left( 1, \frac{\pi}{4} \right)\] and tangent at any point of which makes an angle tan−1 \[\left( \frac{y}{x} - \cos^2 \frac{y}{x} \right)\] with x-axis.
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.
The equation of the curve whose slope is given by \[\frac{dy}{dx} = \frac{2y}{x}; x > 0, y > 0\] and which passes through the point (1, 1) is
Which of the following is the integrating factor of (x log x) \[\frac{dy}{dx} + y\] = 2 log 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.
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.
`x^2 dy/dx = x^2 +xy - y^2`
State whether the following is True or False:
The integrating factor of the differential equation `dy/dx - y = x` is e-x
The function y = ex is solution ______ of differential equation
Solve the differential equation
`x + y dy/dx` = x2 + y2
