हिंदी

The Integrating Factor of the Differential Equation X D Y D X − Y = 2 X 2 - Mathematics

Advertisements
Advertisements

प्रश्न

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

विकल्प

  • e−x

  • ey

  • \[\frac{1}{x}\]

  • x

MCQ

उत्तर

\[\frac{1}{x}\]
 
We have,
\[x\frac{dy}{dx} - y = 2 x^2 \]
\[ \Rightarrow \frac{dy}{dx} - \frac{1}{x}y = 2x\]
\[\text{ Comparing with }\frac{dy}{dx} + Py = Q,\text{ we get }\]
\[P = - \frac{1}{x} \]
\[Q = 2x\]
Now,
\[I . F . = e^{- \int\frac{1}{x}dy} \]
\[ = e^{- \log\left| x \right|} \]
\[ = e^{log\left| \frac{1}{x} \right|} \]
\[ = \frac{1}{x}\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 22: Differential Equations - MCQ [पृष्ठ १४३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 22 Differential Equations
MCQ | Q 50 | पृष्ठ १४३

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

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


Show that Ax2 + By2 = 1 is a solution of the differential equation x \[\left\{ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right\} = y\frac{dy}{dx}\]

 


For the following differential equation verify that the accompanying function is a solution:

Differential equation Function
\[x\frac{dy}{dx} = y\]
y = ax

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

Function y = ex + 1


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

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

C' (x) = 2 + 0.15 x ; C(0) = 100


\[\frac{dy}{dx} = \frac{1 + y^2}{y^3}\]

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

(ey + 1) cos x dx + ey sin x dy = 0


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

 


Solve the differential equation \[\frac{dy}{dx} = \frac{2x\left( \log x + 1 \right)}{\sin y + y \cos y}\], given that y = 0, when x = 1.


Find the solution of the differential equation cos y dy + cos x sin y dx = 0 given that y = \[\frac{\pi}{2}\], when x = \[\frac{\pi}{2}\] 

 


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

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

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

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:-
\[x\frac{dy}{dx} - y = \log x, y\left( 1 \right) = 0\]


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:
\[x\frac{dy}{dx} + y = x \cos x + \sin x, y\left( \frac{\pi}{2} \right) = 1\]


Solve the following initial value problem:-
\[\tan x\left( \frac{dy}{dx} \right) = 2x\tan x + x^2 - y; \tan x \neq 0\] given that y = 0 when \[x = \frac{\pi}{2}\]


If the marginal cost of manufacturing a certain item is given by C' (x) = \[\frac{dC}{dx}\] = 2 + 0.15 x. Find the total cost function C (x), given that C (0) = 100.

 

Which of the following transformations reduce the differential equation \[\frac{dz}{dx} + \frac{z}{x}\log z = \frac{z}{x^2} \left( \log z \right)^2\] into the form \[\frac{du}{dx} + P\left( x \right) u = Q\left( x \right)\]


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


Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of x-axis.


Choose the correct option from the given alternatives:

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


Determine the order and degree of the following differential equations.

Solution D.E.
ax2 + by2 = 5 `xy(d^2y)/dx^2+ x(dy/dx)^2 = y dy/dx`

Solve the following differential equation.

`dy /dx +(x-2 y)/ (2x- y)= 0`


Solve the following differential equation.

`x^2 dy/dx = x^2 +xy - y^2`


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.


Solve the differential equation:

dr = a r dθ − θ dr


Solve:

(x + y) dy = a2 dx


The solution of differential equation `x^2 ("d"^2y)/("d"x^2)` = 1 is ______


Solve the following differential equation 

sec2 x tan y dx + sec2 y tan x dy = 0

Solution: sec2 x tan y dx + sec2 y tan x dy = 0

∴ `(sec^2x)/tanx  "d"x + square` = 0

Integrating, we get

`square + int (sec^2y)/tany  "d"y` = log c

Each of these integral is of the type

`int ("f'"(x))/("f"(x))  "d"x` = log |f(x)| + log c

∴ the general solution is

`square + log |tan y|` = log c

∴ log |tan x . tan y| = log c

`square`

This is the general solution.


lf the straight lines `ax + by + p` = 0 and `x cos alpha + y sin alpha = p` are inclined at an angle π/4 and concurrent with the straight line `x sin alpha - y cos alpha` = 0, then the value of `a^2 + b^2` is


Solve the differential equation

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×