मराठी

√ D 2 Y D X 2 = √ D Y D X

Advertisements
Advertisements

प्रश्न

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

उत्तर

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

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

Taking cubes of both the sides, we get

\[ \Rightarrow \frac{d^2 y}{d x^2} = \left( \frac{dy}{dx} \right)^\frac{3}{2} \]

Squaring both the sides, we get

\[ \Rightarrow \left( \frac{d^2 y}{d x^2} \right)^2 = \left( \frac{dy}{dx} \right)^3 \]

\[ \Rightarrow \left( \frac{d^2 y}{d x^2} \right)^2 - \left( \frac{dy}{dx} \right)^3 = 0\]

In this differential equation, the order of the highest order derivative is 2 and its power is 2. So, it is a differential equation of order 2 and degree 2.

Thus, it is a non-linear differential equation, as its degree is 2, which is greater than 1.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 21: Differential Equations - Exercise 22.01 [पृष्ठ ५]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 21 Differential Equations
Exercise 22.01 | Q 6 | पृष्ठ ५

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

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


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


Hence, the given function is the solution to the given differential equation. \[\frac{c - x}{1 + cx}\] is a solution of the differential equation \[(1+x^2)\frac{dy}{dx}+(1+y^2)=0\].


Verify that y = − x − 1 is a solution of the differential equation (y − x) dy − (y2 − x2) dx = 0.


Verify that y = log \[\left( x + \sqrt{x^2 + a^2} \right)^2\]  satisfies the differential equation \[\left( a^2 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = 0\]


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

\[\frac{dy}{dx} = \log x\]

\[\frac{1}{x}\frac{dy}{dx} = \tan^{- 1} x, x \neq 0\]

(sin x + cos x) dy + (cos x − sin x) dx = 0


\[x\frac{dy}{dx} + 1 = 0 ; y \left( - 1 \right) = 0\]

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

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

(y + xy) dx + (x − xy2) dy = 0


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

Solve the differential equation \[x\frac{dy}{dx} + \cot y = 0\] given that \[y = \frac{\pi}{4}\], when \[x=\sqrt{2}\]


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

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

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

Find the equation of the curve which passes through the origin and has the slope x + 3y− 1 at any point (x, y) on it.


Find the equation of the curve which passes through the point (1, 2) and the distance between the foot of the ordinate of the point of contact and the point of intersection of the tangent with x-axis is twice the abscissa of the point of contact.


Find the equation of the curve that passes through the point (0, a) and is such that at any point (x, y) on it, the product of its slope and the ordinate is equal to the abscissa.


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


The solution of the differential equation \[\frac{dy}{dx} = \frac{ax + g}{by + f}\] represents a circle when


The differential equation of the ellipse \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = C\] is


Find the coordinates of the centre, foci and equation of directrix of the hyperbola x2 – 3y2 – 4x = 8.


Solve the following differential equation.

`(dθ)/dt  = − k (θ − θ_0)`


For  the following differential equation find the particular solution.

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

when  y = 1, x = 0


Solve the following differential equation.

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


Solve the following differential equation.

`dy/dx + y` = 3


Solve the following differential equation.

y dx + (x - y2 ) dy = 0


x2y dx – (x3 + y3) dy = 0


Select and write the correct alternative from the given option for the question

The differential equation of y = Ae5x + Be–5x is


Solve: `("d"y)/("d"x) + 2/xy` = x2 


Solve the following differential equation y log y = `(log  y - x) ("d"y)/("d"x)`


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.


An appropriate substitution to solve the differential equation `"dx"/"dy" = (x^2 log(x/y) - x^2)/(xy log(x/y))` is ______.


Solve the differential equation `"dy"/"dx" + 2xy` = y


Solve the differential equation

`x + y dy/dx` = x2 + y2


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×