हिंदी

Show that the Differential Equation of Which Y = 2(X2 − 1) + C E − X 2 is a Solution, is D Y D X + 2 X Y = 4 X 3 - Mathematics

Advertisements
Advertisements

प्रश्न

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

Advertisements

उत्तर

The given equation is \[y = 2\left( x^2 - 1 \right) + c e^{- x^2}\]                                       ...(1)
where  c  is a parameter.
As this equation has one arbitrary constant, we shall get a differential equation of first order.
Differentiating equation (1) with respect to x, we get
\[\frac{dy}{dx} = 2\left( 2x \right) + c e^{- x^2} ( - 2x)\]
\[ \Rightarrow \frac{dy}{dx} = 4x - 2xc e^{- x^2} . . . \left( 2 \right)\]
From (1) and (2), we get
\[\Rightarrow \frac{dy}{dx} = 4x - 2xy + 4 x^3 - 4x\]
\[ \Rightarrow \frac{dy}{dx} + 2xy = 4 x^3\]
Hence,
\[y = 2\left( x^2 - 1 \right) + c e^{- x^2}\]  is the solution to the differential equation \[\frac{dy}{dx} + 2xy = 4 x^3\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 22: Differential Equations - Exercise 22.02 [पृष्ठ १७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 22 Differential Equations
Exercise 22.02 | Q 13 | पृष्ठ १७

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

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

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

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


Show that y = AeBx is a solution of the differential equation

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

Verify that y2 = 4ax is a solution of the differential equation y = x \[\frac{dy}{dx} + a\frac{dx}{dy}\]


Verify that y2 = 4a (x + a) is a solution of the differential equations
\[y\left\{ 1 - \left( \frac{dy}{dx} \right)^2 \right\} = 2x\frac{dy}{dx}\]


\[\frac{dy}{dx} = x^2 + x - \frac{1}{x}, x \neq 0\]

\[\frac{dy}{dx} = x^5 + x^2 - \frac{2}{x}, x \neq 0\]

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

\[\sqrt{1 - x^4} dy = x\ dx\]

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

Solve the following differential equation:
\[\left( 1 + y^2 \right) \tan^{- 1} xdx + 2y\left( 1 + x^2 \right)dy = 0\]


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

\[\frac{dy}{dx} = 2 e^x y^3 , y\left( 0 \right) = \frac{1}{2}\]

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

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

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


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

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

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:-

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


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 \[x = \frac{\pi}{2}\]


Radium decomposes at a rate proportional to the quantity of radium present. It is found that in 25 years, approximately 1.1% of a certain quantity of radium has decomposed. Determine approximately how long it will take for one-half of the original amount of  radium to decompose?


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


The integrating factor of the differential equation (x log x)
\[\frac{dy}{dx} + y = 2 \log x\], is given by


The differential equation
\[\frac{dy}{dx} + Py = Q y^n , n > 2\] can be reduced to linear form by substituting


Determine the order and degree of the following differential equations.

Solution D.E
y = aex + be−x `(d^2y)/dx^2= 1`

For each of the following differential equations find the particular solution.

(x − y2 x) dx − (y + x2 y) dy = 0, when x = 2, y = 0


Solve the following differential equation.

`dy /dx +(x-2 y)/ (2x- y)= 0`


Solve the following differential equation.

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


Choose the correct alternative.

The solution of `x dy/dx = y` log y is


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

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


Solve the differential equation `("d"y)/("d"x) + y` = e−x 


Solve the following differential equation `("d"y)/("d"x)` = cos(x + y)

Solution: `("d"y)/("d"x)` = cos(x + y)    ......(1)

Put `square`

∴ `1 + ("d"y)/("d"x) = "dv"/("d"x)`

∴ `("d"y)/("d"x) = "dv"/("d"x) - 1`

∴ (1) becomes `"dv"/("d"x) - 1` = cos v

∴ `"dv"/("d"x)` = 1 + cos v

∴ `square` dv = dx

Integrating, we get

`int 1/(1 + cos "v")  "d"v = int  "d"x`

∴ `int 1/(2cos^2 ("v"/2))  "dv" = int  "d"x`

∴ `1/2 int square  "dv" = int  "d"x`

∴ `1/2* (tan("v"/2))/(1/2)` = x + c

∴ `square` = x + c


Given that `"dy"/"dx"` = yex and x = 0, y = e. Find the value of y when x = 1.


The integrating factor of the differential equation `"dy"/"dx" (x log x) + y` = 2logx is ______.


If `y = log_2 log_2(x)` then `(dy)/(dx)` =


Solve the differential equation

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×