मराठी

D Y D X = 1 − X + Y − X Y

Advertisements
Advertisements

प्रश्न

\[\frac{dy}{dx} = 1 - x + y - xy\]
Advertisements

उत्तर

We have, 
\[\frac{dy}{dx} = 1 - x + y - xy\]
\[ \Rightarrow \frac{dy}{dx} = 1 + y - x\left( 1 + y \right)\]
\[ \Rightarrow \frac{dy}{dx} = \left( 1 + y \right)\left( 1 - x \right)\]
\[ \Rightarrow \frac{1}{1 + y}dy = \left( 1 - x \right) dx\]
Integrating both sides, we get
\[\int\frac{1}{1 + y}dy = \int\left( 1 - x \right) dx\]
\[ \Rightarrow \log \left| 1 + y \right| = x - \frac{x^2}{2} + C\]
\[\text{ Hence, }\log \left| 1 + y \right| = x - \frac{x^2}{2} +\text{ C is the required solution }.\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 21: Differential Equations - Exercise 22.07 [पृष्ठ ५५]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 21 Differential Equations
Exercise 22.07 | Q 30 | पृष्ठ ५५

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

\[\frac{d^2 y}{d x^2} + 4y = 0\]

Show that the function y = A cos 2x − B sin 2x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + 4y = 0\].


Differential equation \[x\frac{dy}{dx} = 1, y\left( 1 \right) = 0\]

Function y = log x


Differential equation \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} = 0, y \left( 0 \right) = 2, y'\left( 0 \right) = 1\]

Function y = ex + 1


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


(sin x + cos x) dy + (cos x − sin x) dx = 0


\[\frac{dy}{dx} - x \sin^2 x = \frac{1}{x \log x}\]

\[\frac{dy}{dx} = x e^x - \frac{5}{2} + \cos^2 x\]

\[\sin\left( \frac{dy}{dx} \right) = k ; y\left( 0 \right) = 1\]

\[\frac{dy}{dx} = \sin^2 y\]

\[y\sqrt{1 + x^2} + x\sqrt{1 + y^2}\frac{dy}{dx} = 0\]

\[x\sqrt{1 - y^2} dx + y\sqrt{1 - x^2} dy = 0\]

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

Solve the following differential equation: 
(xy2 + 2x) dx + (x2 y + 2y) dy = 0


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

 


\[\frac{dr}{dt} = - rt, r\left( 0 \right) = r_0\]

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

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

Find the particular solution of the differential equation \[\frac{dy}{dx} = \frac{xy}{x^2 + y^2}\] given that y = 1 when x = 0.

 


Solve the following initial value problem:-

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


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


The differential equation satisfied by ax2 + by2 = 1 is


What is integrating factor of \[\frac{dy}{dx}\] + y sec x = tan x?


The integrating factor of the differential equation \[x\frac{dy}{dx} - y = 2 x^2\]


The differential equation `y dy/dx + x = 0` represents family of ______.


In the following example, verify that the given function is a solution of the corresponding differential equation.

Solution D.E.
y = xn `x^2(d^2y)/dx^2 - n xx (xdy)/dx + ny =0`

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.

`dy/dx + y = e ^-x`


Solve the following differential equation.

`(x + y) dy/dx = 1`


Solve the following differential equation.

`(x + a) dy/dx = – y + a`


y2 dx + (xy + x2)dy = 0


y dx – x dy + log x dx = 0


Solve the differential equation (x2 – yx2)dy + (y2 + xy2)dx = 0


Choose the correct alternative:

Solution of the equation `x("d"y)/("d"x)` = y log y is


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 


Solve `x^2 "dy"/"dx" - xy = 1 + cos(y/x)`, x ≠ 0 and x = 1, y = `pi/2`


Solve the differential equation `"dy"/"dx"` = 1 + x + y2 + xy2, when y = 0, x = 0.


Solution of `x("d"y)/("d"x) = y + x tan  y/x` is `sin(y/x)` = cx


Solve the differential equation

`y (dy)/(dx) + x` = 0


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×