मराठी

Solve the Following Differential Equation: Y E X Y D X = ( X E X Y + Y 2 ) D Y , Y ≠ 0

Advertisements
Advertisements

प्रश्न

Solve the following differential equation:
\[y e^\frac{x}{y} dx = \left( x e^\frac{x}{y} + y^2 \right)dy, y \neq 0\]

 

Advertisements

उत्तर

\[y e^\frac{x}{y} dx = \left( x e^\frac{x}{y} + y^2 \right)dy\]
\[ \Rightarrow y e^\frac{x}{y} dx = x e^\frac{x}{y} dy + y^2 dy\]
\[ \Rightarrow y e^\frac{x}{y} dx - x e^\frac{x}{y} dy = y^2 dy\]
\[ \Rightarrow \left( ydx - xdy \right) e^\frac{x}{y} = y^2 dy\]
\[ \Rightarrow \frac{\left( ydx - xdy \right)}{y^2} e^\frac{x}{y} = dy\]
\[ \Rightarrow e^\frac{x}{y} d\left( \frac{x}{y} \right) = dy\]
\[ \Rightarrow \int e^\frac{x}{y} d\left( \frac{x}{y} \right) = \int dy\]
\[ \Rightarrow e^\frac{x}{y} = y + C\]

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 38.3 | पृष्ठ ५५

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

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

\[x^2 \left( \frac{d^2 y}{d x^2} \right)^3 + y \left( \frac{dy}{dx} \right)^4 + y^4 = 0\]

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


Show that y = ex (A cos x + B sin x) is the solution of the differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + 2y = 0\]


Verify that \[y = ce^{tan^{- 1}} x\]  is a solution of the differential equation \[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + \left( 2x - 1 \right)\frac{dy}{dx} = 0\]


Verify that \[y = e^{m \cos^{- 1} x}\] satisfies the differential equation \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} - m^2 y = 0\]


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


\[\frac{dy}{dx} = x^2 + x - \frac{1}{x}, x \neq 0\]

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


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

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

\[5\frac{dy}{dx} = e^x y^4\]

Solve the differential equation \[\frac{dy}{dx} = e^{x + y} + x^2 e^y\].

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

Solve the following differential equation:
\[\left( 1 + y^2 \right) \tan^{- 1} xdx + 2y\left( 1 + x^2 \right)dy = 0\]


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

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).


In a culture the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present.


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

\[x^2 \frac{dy}{dx} = x^2 + xy + y^2 \]


\[\frac{dy}{dx} = \frac{y}{x} + \sin\left( \frac{y}{x} \right)\]

 

Solve the following initial value problem:-
\[x\frac{dy}{dx} - y = \log x, y\left( 1 \right) = 0\]


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


Solve the following initial value problem:-

\[\frac{dy}{dx} + 2y \tan x = \sin x; y = 0\text{ when }x = \frac{\pi}{3}\]


Find the equation of the curve that passes through the point (0, a) and is such that at any point (x, y) on it, the product of its slope and the ordinate is equal to the abscissa.


The differential equation
\[\frac{dy}{dx} + Py = Q y^n , n > 2\] can be reduced to linear form by substituting


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


Choose the correct option from the given alternatives:

The solution of `1/"x" * "dy"/"dx" = tan^-1 "x"` is


A solution of a differential equation which can be obtained from the general solution by giving particular values to the arbitrary constants is called ___________ solution.


y2 dx + (xy + x2)dy = 0


Solve: `("d"y)/("d"x) + 2/xy` = x2 


Solve the following differential equation y log y = `(log  y - x) ("d"y)/("d"x)`


For the differential equation, find the particular solution

`("d"y)/("d"x)` = (4x +y + 1), when y = 1, x = 0


The function y = cx is the solution of differential equation `("d"y)/("d"x) = y/x`


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


Solve the following differential equation `("d"y)/("d"x)` = cos(x + y)

Solution: `("d"y)/("d"x)` = cos(x + y)    ......(1)

Put `square`

∴ `1 + ("d"y)/("d"x) = "dv"/("d"x)`

∴ `("d"y)/("d"x) = "dv"/("d"x) - 1`

∴ (1) becomes `"dv"/("d"x) - 1` = cos v

∴ `"dv"/("d"x)` = 1 + cos v

∴ `square` dv = dx

Integrating, we get

`int 1/(1 + cos "v")  "d"v = int  "d"x`

∴ `int 1/(2cos^2 ("v"/2))  "dv" = int  "d"x`

∴ `1/2 int square  "dv" = int  "d"x`

∴ `1/2* (tan("v"/2))/(1/2)` = x + c

∴ `square` = x + c


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


An appropriate substitution to solve the differential equation `"dx"/"dy" = (x^2 log(x/y) - x^2)/(xy log(x/y))` is ______.


What distinguishes an ordinary differential equation from other types of differential equations?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×