मराठी

Cos Y D Y D X = E X , Y ( 0 ) = π 2

Advertisements
Advertisements

प्रश्न

\[\cos y\frac{dy}{dx} = e^x , y\left( 0 \right) = \frac{\pi}{2}\]
बेरीज
Advertisements

उत्तर

\[\cos y\frac{dy}{dx} = e^x , y\left( 0 \right) = \frac{\pi}{2}\]
\[ \Rightarrow \cos y\ dy = e^x dx\]
Integrating both sides, we get
\[\int\cos y\ dy = \int e^x dx\]
\[ \Rightarrow \sin y = e^x + C . . . . . (1)\]
\[\text{ We know that at }x = 0, y = \frac{\pi}{2} . \]
Substituting the values of x and y in (1), we get
\[1 = 1 + C\]
\[ \Rightarrow C = 0\]
Substituting the value of C in (1), we get
\[\sin y = e^x \]
\[ \Rightarrow y = \sin^{- 1} \left( e^x \right)\]
\[\text{ Hence, }y = \sin^{- 1} \left( e^x \right)\text{ is the required solution }.\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 21: Differential Equations - Exercise 22.07 [पृष्ठ ५६]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 21 Differential Equations
Exercise 22.07 | Q 45.4 | पृष्ठ ५६

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

\[x + \left( \frac{dy}{dx} \right) = \sqrt{1 + \left( \frac{dy}{dx} \right)^2}\]

Form the differential equation representing the family of ellipses having centre at the origin and foci on x-axis.


Show that y = AeBx is a solution of the differential equation

\[\frac{d^2 y}{d x^2} = \frac{1}{y} \left( \frac{dy}{dx} \right)^2\]

Verify that y = cx + 2c2 is a solution of the differential equation 

\[2 \left( \frac{dy}{dx} \right)^2 + x\frac{dy}{dx} - y = 0\].

Differential equation \[\frac{d^2 y}{d x^2} + y = 0, y \left( 0 \right) = 0, y' \left( 0 \right) = 1\] Function y = sin x


Differential equation \[\frac{dy}{dx} + y = 2, y \left( 0 \right) = 3\] Function y = e−x + 2


\[\frac{dy}{dx} = \log x\]

\[\frac{1}{x}\frac{dy}{dx} = \tan^{- 1} x, x \neq 0\]

\[\frac{dy}{dx} = \cos^3 x \sin^2 x + x\sqrt{2x + 1}\]

\[\left( 1 + x^2 \right)\frac{dy}{dx} - x = 2 \tan^{- 1} x\]

\[\cos x \cos y\frac{dy}{dx} = - \sin x \sin y\]

\[\frac{dy}{dx} + \frac{\cos x \sin y}{\cos y} = 0\]

\[\left( x - 1 \right)\frac{dy}{dx} = 2 x^3 y\]

Solve the following differential equation:
\[y\left( 1 - x^2 \right)\frac{dy}{dx} = x\left( 1 + y^2 \right)\]

 


\[\frac{dy}{dx} = y \tan 2x, y\left( 0 \right) = 2\] 

\[\frac{dy}{dx} = 2 e^x y^3 , y\left( 0 \right) = \frac{1}{2}\]

Find the particular solution of the differential equation
(1 – y2) (1 + log x) dx + 2xy dy = 0, given that y = 0 when x = 1.


(y2 − 2xy) dx = (x2 − 2xy) dy


\[\frac{dy}{dx} = \frac{y}{x} - \sqrt{\frac{y^2}{x^2} - 1}\]

\[\left[ x\sqrt{x^2 + y^2} - y^2 \right] dx + xy\ dy = 0\]

Solve the following initial value problem:
\[x\frac{dy}{dx} + y = x \cos x + \sin x, y\left( \frac{\pi}{2} \right) = 1\]


Solve the following initial value problem:-

\[\frac{dy}{dx} - 3y \cot x = \sin 2x; y = 2\text{ when }x = \frac{\pi}{2}\]


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 origin and has the slope x + 3y− 1 at any point (x, y) on it.


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.


Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y\] sin x = 1, is


The integrating factor of the differential equation \[\left( 1 - y^2 \right)\frac{dx}{dy} + yx = ay\left( - 1 < y < 1 \right)\] is ______.


Form the differential equation of the family of circles having centre on y-axis and radius 3 unit.


Find the differential equation whose general solution is

x3 + y3 = 35ax.


Solve the following differential equation.

`xy  dy/dx = x^2 + 2y^2`


Solve the following differential equation.

`dy/dx + 2xy = x`


Solve the following differential equation.

`(x + a) dy/dx = – y + a`


Solve:

(x + y) dy = a2 dx


State whether the following statement is True or False:

The integrating factor of the differential equation `("d"y)/("d"x) - y` = x is e–x 


Verify y = log x + c is the solution of differential equation `x ("d"^2y)/("d"x^2) + ("d"y)/("d"x)` = 0


Solve: `("d"y)/("d"x) = cos(x + y) + sin(x + y)`. [Hint: Substitute x + y = z]


If `y = log_2 log_2(x)` then `(dy)/(dx)` =


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


`d/(dx)(tan^-1  (sqrt(1 + x^2) - 1)/x)` is equal to:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×