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A solution of differential equation which can be obtained from the general solution by giving particular values to the arbitrary constant is called ______ solution - Mathematics and Statistics

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प्रश्न

A solution of differential equation which can be obtained from the general solution by giving particular values to the arbitrary constant is called ______ solution

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उत्तर

particular

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अध्याय 1.8: Differential Equation and Applications - Q.2

संबंधित प्रश्न

\[\frac{d^2 y}{d x^2} + 4y = 0\]

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


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


\[\sqrt{1 - x^4} dy = x\ dx\]

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

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

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

(ey + 1) cos x dx + ey sin x dy = 0


tan y \[\frac{dy}{dx}\] = sin (x + y) + sin (x − y) 

 


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

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

\[2x\frac{dy}{dx} = 5y, y\left( 1 \right) = 1\]

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

\[2\left( y + 3 \right) - xy\frac{dy}{dx} = 0\], y(1) = −2

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

Solve the following initial value problem:-

\[\frac{dy}{dx} - 3y \cot x = \sin 2x; y = 2\text{ when }x = \frac{\pi}{2}\]


The decay rate of radium at any time t is proportional to its mass at that time. Find the time when the mass will be halved of its initial mass.


Find the equation of the curve such that the portion of the x-axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2).


At every point on a curve the slope is the sum of the abscissa and the product of the ordinate and the abscissa, and the curve passes through (0, 1). Find the equation of the curve.


The slope of a curve at each of its points is equal to the square of the abscissa of the point. Find the particular curve through the point (−1, 1).


The equation of the curve whose slope is given by \[\frac{dy}{dx} = \frac{2y}{x}; x > 0, y > 0\] and which passes through the point (1, 1) is


Which of the following is the integrating factor of (x log x) \[\frac{dy}{dx} + y\] = 2 log x?


In the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:-

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


For  the following differential equation find the particular solution.

`dy/ dx = (4x + y + 1),

when  y = 1, x = 0


Solve

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


`xy dy/dx  = x^2 + 2y^2`


Solve the following differential equation

`y log y ("d"x)/("d"y) + x` = log y


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


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