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

The function y = cx is the solution of differential equation dddydx=yx

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प्रश्न

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

पर्याय

  • True

  • False

MCQ
चूक किंवा बरोबर
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उत्तर

This statement is True.

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 1.8: Differential Equation and Applications - Q.3

संबंधित प्रश्‍न

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


Show that y = ax3 + bx2 + c is a solution of the differential equation \[\frac{d^3 y}{d x^3} = 6a\].

 


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

Differential equation Function
\[x\frac{dy}{dx} + y = y^2\]
\[y = \frac{a}{x + a}\]

Differential equation \[\frac{dy}{dx} + y = 2, y \left( 0 \right) = 3\] Function y = e−x + 2


Differential equation \[\frac{d^2 y}{d x^2} - 3\frac{dy}{dx} + 2y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 3\] Function y = ex + e2x


\[\frac{dy}{dx} + 2x = e^{3x}\]

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

C' (x) = 2 + 0.15 x ; C(0) = 100


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

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

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


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

 


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

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).


If y(x) is a solution of the different equation \[\left( \frac{2 + \sin x}{1 + y} \right)\frac{dy}{dx} = - \cos x\] and y(0) = 1, then find the value of y(π/2).


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

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

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


The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also find the time necessary for the number of bacteria to be 10 times the number of initial present.


The solution of the differential equation y1 y3 = y22 is


Verify that the function y = e−3x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + \frac{dy}{dx} - 6y = 0.\]


Choose the correct option from the given alternatives:

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


Form the differential equation from the relation x2 + 4y2 = 4b2


Solve the following differential equation.

x2y dx − (x3 + y3) dy = 0


Solve the following differential equation.

(x2 − y2 ) dx + 2xy dy = 0


The solution of `dy/dx + x^2/y^2 = 0` is ______


`xy dy/dx  = x^2 + 2y^2`


 `dy/dx = log x`


Given that `"dy"/"dx"` = yex and x = 0, y = e. Find the value of y when x = 1.


Solve the differential equation `dy/dx + xy = xy^2` and find the particular solution when y = 4, x = 1.


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