Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\frac{dy}{dx} = y \tan x, y\left( 0 \right) = 1\]
\[ \Rightarrow \frac{1}{y}dy = \tan x dx\]
Integrating both sides, we get
\[\int\frac{1}{y}dy = \int\tan x dx\]
\[ \Rightarrow \log \left| y \right| = \log \left| \sec x \right| + C . . . . . (1)\]
We know that at x = 0 and y = 1 .
Substituting the values of x and y in (1), we get
\[\log \left| 1 \right| = \log \left| 1 \right| + C\]
\[ \Rightarrow C = 0\]
Substituting the value of C in (1), we get
\[\log \left| y \right| = \log \left| \sec x \right| + 0\]
\[ \Rightarrow y = \sec x\]
\[\text{ Hence, }y = \sec x,\text{ where }x \in \left( \frac{- \pi}{2}, \frac{\pi}{2} \right),\text{ is the required solution .}\]
APPEARS IN
संबंधित प्रश्न
Verify that y = 4 sin 3x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + 9y = 0\]
Differential equation \[\frac{dy}{dx} + y = 2, y \left( 0 \right) = 3\] Function y = e−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 dy = (y − 1) (x + 1) dx
dy + (x + 1) (y + 1) dx = 0
In a bank principal increases at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0.5 = 1.648).
\[x^2 \frac{dy}{dx} = x^2 + xy + y^2 \]
Solve the following initial value problem:-
\[\left( 1 + y^2 \right) dx + \left( x - e^{- \tan^{- 1} y} \right) dx = 0, y\left( 0 \right) = 0\]
Solve the following initial value problem:
\[x\frac{dy}{dx} + y = x \cos x + \sin x, y\left( \frac{\pi}{2} \right) = 1\]
A population grows at the rate of 5% per year. How long does it take for the population to double?
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.
Find the equation of the curve which passes through the point (2, 2) and satisfies the differential equation
\[y - x\frac{dy}{dx} = y^2 + \frac{dy}{dx}\]
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 integrating factor of the differential equation (x log x)
\[\frac{dy}{dx} + y = 2 \log x\], is given by
Find the coordinates of the centre, foci and equation of directrix of the hyperbola x2 – 3y2 – 4x = 8.
Find the differential equation whose general solution is
x3 + y3 = 35ax.
Solve the following differential equation.
`(x + y) dy/dx = 1`
The differential equation of `y = k_1e^x+ k_2 e^-x` is ______.
y dx – x dy + log x dx = 0
Solve `("d"y)/("d"x) = (x + y + 1)/(x + y - 1)` when x = `2/3`, y = `1/3`
Solve the following differential equation
`yx ("d"y)/("d"x)` = x2 + 2y2
Solve the following differential equation y log y = `(log y - x) ("d"y)/("d"x)`
A solution of differential equation which can be obtained from the general solution by giving particular values to the arbitrary constant is called ______ solution
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`
