English

Verify that Y = Cx + 2c2 is a Solution of the Differential Equation 2 ( D Y D X ) 2 + X D Y D X − Y = 0 . - Mathematics

Advertisements
Advertisements

Question

Verify that y = cx + 2c2 is a solution of the differential equation 

\[2 \left( \frac{dy}{dx} \right)^2 + x\frac{dy}{dx} - y = 0\].
Sum

Solution

We have,
\[y = cx + 2 c^2..............(1)\]
Differentiating both sides of (1) with respect to x, we get

\[\frac{dy}{dx} = c...........(2)\]
Now,
\[2 \left( \frac{dy}{dx} \right)^2 + x\frac{dy}{dx} - y\]
\[ = 2 c^2 + cx - cx - 2 c^2 = 0 ...........\left[\text{Using }\left( 1 \right)\text{ and }\left( 2 \right) \right]\]
\[ \Rightarrow 2 \left( \frac{dy}{dx} \right)^2 + x\frac{dy}{dx} - y = 0\]
Hence, the given function is the solution to the given differential equation.
shaalaa.com
  Is there an error in this question or solution?
Chapter 22: Differential Equations - Exercise 22.03 [Page 25]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 22 Differential Equations
Exercise 22.03 | Q 13 | Page 25

Video TutorialsVIEW ALL [2]

RELATED QUESTIONS

Find the differential equation of all the parabolas with latus rectum '4a' and whose axes are parallel to x-axis.


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

Differential equation Function
\[x\frac{dy}{dx} = y\]
y = ax

\[\sin^4 x\frac{dy}{dx} = \cos x\]

\[\sqrt{a + x} dy + x\ dx = 0\]

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

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

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

(ey + 1) cos x dx + ey sin x dy = 0


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

\[\sqrt{1 + x^2} dy + \sqrt{1 + y^2} dx = 0\]

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

 


\[x\sqrt{1 - y^2} dx + y\sqrt{1 - x^2} dy = 0\]

dy + (x + 1) (y + 1) dx = 0


Solve the following differential equation: 
(xy2 + 2x) dx + (x2 y + 2y) dy = 0


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

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

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

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

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

Solve the following initial value problem:-

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


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.


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


Which of the following is the integrating factor of (x log x) \[\frac{dy}{dx} + y\] = 2 log x?


The integrating factor of the differential equation \[\left( 1 - y^2 \right)\frac{dx}{dy} + yx = ay\left( - 1 < y < 1 \right)\] is ______.


Find the particular solution of the differential equation `"dy"/"dx" = "xy"/("x"^2+"y"^2),`given that y = 1 when x = 0


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

Solution D.E.
xy = log y + k y' (1 - xy) = y2

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`

For  the following differential equation find the particular solution.

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

when  y = 1, x = 0


Solve the following differential equation.

x2y dx − (x3 + y3 ) dy = 0


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.


x2y dx – (x3 + y3) dy = 0


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

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


Solve the differential equation xdx + 2ydy = 0


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


Solve: ydx – xdy = x2ydx.


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


`d/(dx)(tan^-1  (sqrt(1 + x^2) - 1)/x)` is equal to:


Solve the differential equation

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×