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

The solution of differential equation x2d2ydx2 = 1 is ______

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प्रश्न

The solution of differential equation `x^2 ("d"^2y)/("d"x^2)` = 1 is ______

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उत्तर

y = 1 – log x

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पाठ 1.8: Differential Equation and Applications - Q.2

संबंधित प्रश्‍न

\[\sqrt{1 + \left( \frac{dy}{dx} \right)^2} = \left( c\frac{d^2 y}{d x^2} \right)^{1/3}\]

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


C' (x) = 2 + 0.15 x ; C(0) = 100


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

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

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

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

Find the particular solution of edy/dx = x + 1, given that y = 3, when x = 0.


x2 dy + y (x + y) dx = 0


Solve the following initial value problem:-

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


Solve the following initial value problem:-

\[\left( 1 + y^2 \right) dx + \left( x - e^{- \tan^{- 1} y} \right) dx = 0, y\left( 0 \right) = 0\]


Find the equation of the curve passing through the point \[\left( 1, \frac{\pi}{4} \right)\]  and tangent at any point of which makes an angle tan−1  \[\left( \frac{y}{x} - \cos^2 \frac{y}{x} \right)\] with x-axis.


The solution of the differential equation y1 y3 = y22 is


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


Which of the following differential equations has y = C1 ex + C2 ex as the general solution?


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


If a + ib = `("x" + "iy")/("x" - "iy"),` prove that `"a"^2 +"b"^2 = 1` and `"b"/"a" = (2"xy")/("x"^2 - "y"^2)`


Choose the correct option from the given alternatives:

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


Solve the following differential equation.

y2 dx + (xy + x2 ) dy = 0


Solve the following differential equation.

`xy  dy/dx = x^2 + 2y^2`


Solve the following differential equation.

y dx + (x - y2 ) dy = 0


Solve the following differential equation.

`dy/dx + 2xy = x`


Choose the correct alternative.

The differential equation of y = `k_1 + k_2/x` is


A solution of a differential equation which can be obtained from the general solution by giving particular values to the arbitrary constants is called ___________ solution.


Solve

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


`xy dy/dx  = x^2 + 2y^2`


Solve the following differential equation

`yx ("d"y)/("d"x)` = x2 + 2y2 


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.


Solve the differential equation `"dy"/"dx"` = 1 + x + y2 + xy2, when y = 0, x = 0.


Solve the differential equation

`y (dy)/(dx) + x` = 0


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