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

The function y = ex is solution ______ of differential equation

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प्रश्न

The function y = ex is solution  ______ of differential equation

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उत्तर

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

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पाठ 1.8: Differential Equation and Applications - Q.2

संबंधित प्रश्‍न

\[\frac{d^3 x}{d t^3} + \frac{d^2 x}{d t^2} + \left( \frac{dx}{dt} \right)^2 = e^t\]

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


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


Show that y = ex (A cos x + B sin x) is the solution of the differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + 2y = 0\]


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

Differential equation Function
\[x + y\frac{dy}{dx} = 0\]
\[y = \pm \sqrt{a^2 - x^2}\]

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

Differential equation Function
\[x^3 \frac{d^2 y}{d x^2} = 1\]
\[y = ax + b + \frac{1}{2x}\]

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


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

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


\[\frac{dy}{dx} = x e^x - \frac{5}{2} + \cos^2 x\]

\[\sin\left( \frac{dy}{dx} \right) = k ; y\left( 0 \right) = 1\]

\[\frac{dy}{dx} = \frac{1 - \cos 2y}{1 + \cos 2y}\]

(1 + x2) dy = xy dx


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

 


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

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

x2 dy + y (x + y) dx = 0


\[\left[ x\sqrt{x^2 + y^2} - y^2 \right] dx + xy\ dy = 0\]

Find the equation of the curve which passes through the point (2, 2) and satisfies the differential equation
\[y - x\frac{dy}{dx} = y^2 + \frac{dy}{dx}\]


A curve is such that the length of the perpendicular from the origin on the tangent at any point P of the curve is equal to the abscissa of P. Prove that the differential equation of the curve is \[y^2 - 2xy\frac{dy}{dx} - x^2 = 0\], and hence find the curve.


Write the differential equation representing the family of straight lines y = Cx + 5, where C is an arbitrary constant.


Solve the following differential equation.

xdx + 2y dx = 0


Solve the following differential equation.

`dy/dx + 2xy = x`


The integrating factor of the differential equation `dy/dx - y = x` is e−x.


 `dy/dx = log x`


Integrating factor of the differential equation `"dy"/"dx" - y` = cos x is ex.


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


Which of the following defines a differential equation?


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