English

D 4 Y D X 4 = { C + ( D Y D X ) 2 } 3 / 2 - Mathematics

Advertisements
Advertisements

Question

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

Solution

\[\frac{d^4 y}{d x^4} = \left[ c + \left( \frac{dy}{dx} \right)^2 \right]^\frac{3}{2} \]
Squaring both sides, we get
\[ \Rightarrow \left( \frac{d^4 y}{d x^4} \right)^2 = \left[ c + \left( \frac{dy}{dx} \right)^2 \right]^3 \]
\[ \Rightarrow \left( \frac{d^4 y}{d x^4} \right)^2 = c^3 + 3 c^2 \left( \frac{dy}{dx} \right)^2 + 3c \left( \frac{dy}{dx} \right)^4 + \left( \frac{dy}{dx} \right)^6\]
In this differential equation, the order of the highest order derivative is 4 and its power is 2. So, it is a differential equation of order 4 and degree 2.
Thus, it is a non-linear differential equation, as its degree is 2, which is greater than 1.

shaalaa.com
  Is there an error in this question or solution?
Chapter 22: Differential Equations - Exercise 22.01 [Page 5]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 22 Differential Equations
Exercise 22.01 | Q 7 | Page 5

Video TutorialsVIEW ALL [2]

RELATED QUESTIONS

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

Show that y = ax3 + bx2 + c is a solution of the differential equation \[\frac{d^3 y}{d x^3} = 6a\].

 


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


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


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

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

C' (x) = 2 + 0.15 x ; C(0) = 100


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

Solve the following differential equation:
\[y\left( 1 - x^2 \right)\frac{dy}{dx} = x\left( 1 + y^2 \right)\]

 


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

Solve the differential equation \[\frac{dy}{dx} = \frac{2x\left( \log x + 1 \right)}{\sin y + y \cos y}\], given that y = 0, when x = 1.


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

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

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

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

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

Solve the following initial value problem:-
\[x\frac{dy}{dx} - y = \log x, y\left( 1 \right) = 0\]


Solve the following initial value problem:-

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


Solve the following initial value problem:-
\[\tan x\left( \frac{dy}{dx} \right) = 2x\tan x + x^2 - y; \tan x \neq 0\] given that y = 0 when \[x = \frac{\pi}{2}\]


A bank pays interest by continuous compounding, that is, by treating the interest rate as the instantaneous rate of change of principal. Suppose in an account interest accrues at 8% per year, compounded continuously. Calculate the percentage increase in such an account over one year.


The tangent at any point (x, y) of a curve makes an angle tan−1(2x + 3y) with x-axis. Find the equation of the curve if it passes through (1, 2).


Find the equation of the curve passing through the point (0, 1) if the slope of the tangent to the curve at each of its point is equal to the sum of the abscissa and the product of the abscissa and the ordinate of the point.


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).


What is integrating factor of \[\frac{dy}{dx}\] + y sec x = tan x?


Choose the correct option from the given alternatives:

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


Find the differential equation whose general solution is

x3 + y3 = 35ax.


Form the differential equation from the relation x2 + 4y2 = 4b2


For  the following differential equation find the particular solution.

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

when  y = 1, x = 0


Solve the following differential equation.

(x2 − y2 ) dx + 2xy dy = 0


Choose the correct alternative.

Bacteria increases at the rate proportional to the number present. If the original number M doubles in 3 hours, then the number of bacteria will be 4M in


Solve `("d"y)/("d"x) = (x + y + 1)/(x + y - 1)` when x = `2/3`, y = `1/3`


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

`x^2  ("d"y)/("d"x)` = x2 + xy − y2 


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


The function y = cx is the solution of differential equation `("d"y)/("d"x) = y/x`


Verify y = `a + b/x` is solution of `x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0

y = `a + b/x`

`(dy)/(dx) = square`

`(d^2y)/(dx^2) = square`

Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`

= `x square + 2 square`

= `square`

Hence y = `a + b/x` is solution of `square`


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.


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×