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महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

State whether the following is True or False: The degree of a differential equation is the power of the highest ordered derivative when all the derivatives are made free from negative - Mathematics and Statistics

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प्रश्न

State whether the following is True or False:

The degree of a differential equation is the power of the highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any.

पर्याय

  • True

  • False

MCQ
चूक किंवा बरोबर
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उत्तर

The degree of a differential equation is the power of the highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any. - True

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Differential Equation and Applications - Miscellaneous Exercise 8 [पृष्ठ १७२]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
पाठ 8 Differential Equation and Applications
Miscellaneous Exercise 8 | Q 3.3 | पृष्ठ १७२

संबंधित प्रश्‍न

Assume that a rain drop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the rain drop.

 

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

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

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


(1 − x2) dy + xy dx = xy2 dx


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

dy + (x + 1) (y + 1) dx = 0


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

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


The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after `t` seconds.


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


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

Solve the following initial value problem:-

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


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 \[x = \frac{\pi}{2}\]


The surface area of a balloon being inflated, changes at a rate proportional to time t. If initially its radius is 1 unit and after 3 seconds it is 2 units, find the radius after time t.


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 curve for which the intercept cut-off by a tangent on x-axis is equal to four times the ordinate of the point of contact.

 

The tangent at any point (x, y) of a curve makes an angle tan−1(2x + 3y) with x-axis. Find the equation of the curve if it passes through (1, 2).


Write the differential equation obtained by eliminating the arbitrary constant C in the equation x2 − y2 = C2.


Write the differential equation obtained eliminating the arbitrary constant C in the equation xy = C2.


The solution of the differential equation y1 y3 = y22 is


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


Solve the following differential equation : \[y^2 dx + \left( x^2 - xy + y^2 \right)dy = 0\] .


Solve the following differential equation : \[\left( \sqrt{1 + x^2 + y^2 + x^2 y^2} \right) dx + xy \ dy = 0\].


For the following differential equation find the particular solution.

`(x + 1) dy/dx − 1 = 2e^(−y)`,

when y = 0, x = 1


Solve the following differential equation.

xdx + 2y dx = 0


Solve the following differential equation.

`dy/dx + 2xy = x`


Choose the correct alternative.

The integrating factor of `dy/dx -  y = e^x `is ex, then its solution is


For the differential equation, find the particular solution (x – y2x) dx – (y + x2y) dy = 0 when x = 2, y = 0


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.


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


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