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In each of the following examples, verify that the given function is a solution of the corresponding differential equation. Solution D.E. y = ex dydx=y - Mathematics and Statistics

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प्रश्न

In each of the following examples, verify that the given function is a solution of the corresponding differential equation.

Solution D.E.
y = ex  `dy/ dx= y`
बेरीज

उत्तर

y = ex

Differentiating w.r.t. x, we get

`dy/dx = e^x`

∴ `dy/dx = y`

∴ Given function is a solution of the given differential equation.

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Differential Equation and Applications - Exercise 8.1 [पृष्ठ १६२]

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बालभारती Mathematics and Statistics 1 (Commerce) [English] 12 Standard HSC Maharashtra State Board
पाठ 8 Differential Equation and Applications
Exercise 8.1 | Q 2.3 | पृष्ठ १६२

संबंधित प्रश्‍न

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

\[\sqrt[3]{\frac{d^2 y}{d x^2}} = \sqrt{\frac{dy}{dx}}\]

\[\frac{d^4 y}{d x^4} = \left\{ c + \left( \frac{dy}{dx} \right)^2 \right\}^{3/2}\]

Show that the differential equation of which y = 2(x2 − 1) + \[c e^{- x^2}\] is a solution, is \[\frac{dy}{dx} + 2xy = 4 x^3\]


\[\frac{dy}{dx} - x \sin^2 x = \frac{1}{x \log x}\]

(1 + x2) dy = xy dx


xy dy = (y − 1) (x + 1) dx


\[\frac{dy}{dx} + \frac{\cos x \sin y}{\cos y} = 0\]

y (1 + ex) dy = (y + 1) ex dx


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

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


In a culture, the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present?


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


Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y\] sin x = 1, is


The differential equation \[x\frac{dy}{dx} - y = x^2\], has the general solution


Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y \sin x = 1\], is


Solve the differential equation sec2y tan x dy + sec2x tan y dx = 0


Find the particular solution of the following differential equation

`("d"y)/("d"x)` = e2y cos x, when x = `pi/6`, y = 0.

Solution: The given D.E. is `("d"y)/("d"x)` = e2y cos x

∴ `1/"e"^(2y)  "d"y` = cos x dx

Integrating, we get

`int square  "d"y` = cos x dx

∴ `("e"^(-2y))/(-2)` = sin x + c1

∴ e–2y = – 2sin x – 2c1

∴ `square` = c, where c = – 2c

This is general solution.

When x = `pi/6`, y = 0, we have

`"e"^0 + 2sin  pi/6` = c

∴ c = `square`

∴ particular solution is `square`


Solve `x^2 "dy"/"dx" - xy = 1 + cos(y/x)`, x ≠ 0 and x = 1, y = `pi/2`


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