Advertisements
Advertisements
प्रश्न
Solve the following differential equation
`y log y ("d"x)/("d"y) + x` = log y
Advertisements
उत्तर
`y log y ("d"x)/("d"y) + x` = log y
∴ `("d"x)/("d"y) + 1/(ylogy) x = 1/y`
The given equation is of the form `("d"x)/("d"y) + "P"x` = Q
where, P =`1/(ylogy)` and Q = `1/y`
∴ I.F. = `"e"^(int^("Pd"y)`
= `"e"^(int^(1/(ylogy) "d"y)`
= `"e"^(log |log y|)`
= log y
∴ Solution of the given equation is
x(I.F.) = `int "Q"("I.F.") "d"y + "c"_1`
∴ x.logy = `int 1/y log y "d"y + "c"_1`
In R. H. S., put log y = t
Differentiating w.r.t. x, we get
`1/y "d"y` = dt
∴ x log y = `int "t" "dt" + "c"_1`
= `"t"^2/2 + "c"_1`
∴ x log y = `(log y)^2/2 + "c"_1`
∴ 2x log y = (log y)2 + c ......[2c1 + c]
APPEARS IN
संबंधित प्रश्न
Assume that a rain drop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the rain drop.
Show that y = ax3 + bx2 + c is a solution of the differential equation \[\frac{d^3 y}{d x^3} = 6a\].
(sin x + cos x) dy + (cos x − sin x) dx = 0
C' (x) = 2 + 0.15 x ; C(0) = 100
xy dy = (y − 1) (x + 1) dx
dy + (x + 1) (y + 1) dx = 0
Solve the following differential equation:
(xy2 + 2x) dx + (x2 y + 2y) dy = 0
Solve the following initial value problem:-
\[\frac{dy}{dx} + y\cot x = 2\cos x, y\left( \frac{\pi}{2} \right) = 0\]
Solve the following initial value problem:-
\[dy = \cos x\left( 2 - y\text{ cosec }x \right)dx\]
A population grows at the rate of 5% per year. How long does it take for the population to double?
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?
The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was 200000 in 1990 and 250000 in 2000, what will be the population in 2010?
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 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).
Find the equation of the curve which passes through the point (3, −4) and has the slope \[\frac{2y}{x}\] at any point (x, y) on it.
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 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 by eliminating the arbitrary constant C in the equation x2 − y2 = C2.
The solution of the differential equation y1 y3 = y22 is
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.
x2y dx – (x3 + y3) dy = 0
If `y = log_2 log_2(x)` then `(dy)/(dx)` =
