मराठी

(X + Y − 1) Dy = (X + Y) Dx - Mathematics

Advertisements
Advertisements

प्रश्न

(x + y − 1) dy = (x + y) dx

बेरीज
Advertisements

उत्तर

We have,

\[\left( x + y - 1 \right)dy = \left( x + y \right)dx\]

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

Putting `x + y = v,` we get

\[ \Rightarrow 1 + \frac{dy}{dx} = \frac{dv}{dx}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{dv}{dx} - 1\]

\[ \therefore \frac{dv}{dx} - 1 = \frac{v}{\left( v - 1 \right)}\]

\[ \Rightarrow \frac{dv}{dx} = \frac{v}{\left( v - 1 \right)} + 1\]

\[ \Rightarrow \frac{dv}{dx} = \frac{v + v - 1}{\left( v - 1 \right)}\]

\[ \Rightarrow \frac{dv}{dx} = \frac{2v - 1}{\left( v - 1 \right)}\]

\[ \Rightarrow \frac{v - 1}{2v - 1} dv = dx\]

Integrating both sides, we get

\[\int\frac{v - 1}{2v + 1} dv = \int dx\]

\[ \Rightarrow \frac{1}{2}\int\frac{2v}{2v - 1}dv - \int\frac{1}{2v - 1}dv = \int dx\]

\[ \Rightarrow \frac{1}{2}\int\frac{2v - 1 + 1}{2v - 1}dv - \int\frac{1}{2v - 1}dv = \int dx\]

\[ \Rightarrow \frac{1}{2}\int dv + \frac{1}{2}\int\frac{1}{2v - 1}dv - \int\frac{1}{2v - 1}dv = \int dx\]

\[ \Rightarrow \frac{1}{2}\int dv - \frac{1}{2}\int\frac{1}{2v - 1}dv = \int dx\]

\[ \Rightarrow \frac{1}{2}v - \frac{1}{4}\log \left| 2v - 1 \right| = x + C\]

\[ \Rightarrow \frac{1}{2}\left( x + y \right) - \frac{1}{4}\log \left| 2x + 2y - 1 \right| = x + C\]

\[ \Rightarrow 2\left( x + y \right) - \log \left| 2x + 2y - 1 \right| = 4x + 4C\]

\[ \Rightarrow 2\left( x + y \right) - 4x - \log \left| 2x + 2y - 1 \right| = 4C \]

\[ \Rightarrow 2\left( y - x \right) - \log \left| 2x + 2y - 1 \right| = k,\text{ where }k = 4C\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 22: Differential Equations - Revision Exercise [पृष्ठ १४६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 22 Differential Equations
Revision Exercise | Q 39 | पृष्ठ १४६

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

Find the particular solution of the differential equation `(1+x^2)dy/dx=(e^(mtan^-1 x)-y)` , give that y=1 when x=0.


Solve the differential equation `dy/dx -y =e^x`


Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y = ex + 1  :  y″ – y′ = 0


Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y = x2 + 2x + C  :  y′ – 2x – 2 = 0


Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y = Ax : xy′ = y (x ≠ 0)


The number of arbitrary constants in the general solution of a differential equation of fourth order are ______.


Find a particular solution of the differential equation`(x + 1) dy/dx = 2e^(-y) - 1`, given that y = 0 when x = 0.


if `y = sin^(-1) (6xsqrt(1-9x^2))`, `1/(3sqrt2) < x < 1/(3sqrt2)` then find `(dy)/(dx)`


The general solution of the differential equation \[\frac{dy}{dx} + y \] cot x = cosec x, is


Solution of the differential equation \[\frac{dy}{dx} + \frac{y}{x}=\sin x\] is


The solution of the differential equation \[2x\frac{dy}{dx} - y = 3\] represents


The solution of the differential equation \[\frac{dy}{dx} + 1 = e^{x + y}\], is


The solution of the differential equation (x2 + 1) \[\frac{dy}{dx}\] + (y2 + 1) = 0, is


The general solution of the differential equation \[\frac{dy}{dx} = e^{x + y}\], is


The solution of the differential equation \[\frac{dy}{dx} = \frac{y}{x} + \frac{\phi\left( \frac{y}{x} \right)}{\phi'\left( \frac{y}{x} \right)}\] is


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


\[\frac{dy}{dx} - y \tan x = e^x \sec x\]


(1 + y + x2 y) dx + (x + x3) dy = 0


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


\[\frac{dy}{dx} + 5y = \cos 4x\]


`(dy)/(dx)+ y tan x = x^n cos x, n ne− 1`


For the following differential equation, find the general solution:- `y log y dx − x dy = 0`


Solve the following differential equation:-

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


Solve the following differential equation:-

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


Solve the following differential equation:-

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


Solve the following differential equation:-

(1 + x2) dy + 2xy dx = cot x dx


Find a particular solution of the following differential equation:- (x + y) dy + (x − y) dx = 0; y = 1 when x = 1


Find the equation of a curve passing through the point (0, 0) and whose differential equation is \[\frac{dy}{dx} = e^x \sin x.\]


Solve the differential equation dy = cosx(2 – y cosecx) dx given that y = 2 when x = `pi/2`


Solution of the differential equation tany sec2xdx + tanx sec2ydy = 0 is ______.


The solution of `("d"y)/("d"x) + y = "e"^-x`, y(0) = 0 is ______.


The general solution of the differential equation `("d"y)/("d"x) = "e"^(x^2/2) + xy` is ______.


The differential equation for which y = acosx + bsinx is a solution, is ______.


Solution of the differential equation `("d"y)/("d"x) + y/x` = sec x is ______.


The general solution of the differential equation (ex + 1) ydy = (y + 1) exdx is ______.


Find the particular solution of the differential equation `x (dy)/(dx) - y = x^2.e^x`, given y(1) = 0.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×