हिंदी

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation: y = ex + 1 : y″ – y′ = 0

Advertisements
Advertisements

प्रश्न

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y = ex + 1  :  y″ – y′ = 0

योग
Advertisements

उत्तर

We have y = ex + 1                  ...(1)

Differentiating (1) w.r.t.x, we get

`y' = d/dx (e^x + 1) = e^x`

and `y” = d/dx (e^x) = e^x`

⇒ y” - y’ = 0

Thus, y = ex + 1 is a solution to the stated differentiating (1) equation.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Equations - Exercise 9.2 [पृष्ठ ३८५]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 9 Differential Equations
Exercise 9.2 | Q 1 | पृष्ठ ३८५

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

The differential equation of the family of curves y=c1ex+c2e-x is......

(a)`(d^2y)/dx^2+y=0`

(b)`(d^2y)/dx^2-y=0`

(c)`(d^2y)/dx^2+1=0`

(d)`(d^2y)/dx^2-1=0`


Find the general solution of the following differential equation : 

`(1+y^2)+(x-e^(tan^(-1)y))dy/dx= 0`


Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y – cos y = x :  (y sin y + cos y + x) y′ = y


Find the particular solution of the differential equation

`tan x * (dy)/(dx) = 2x tan x + x^2 - y`; `(tan x != 0)` given that y = 0 when `x = pi/2`


The solution of the differential equation \[\frac{dy}{dx} = 1 + x + y^2 + x y^2 , y\left( 0 \right) = 0\] is


The solution of x2 + y \[\frac{dy}{dx}\]= 4, is


The solution of the differential equation \[\frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2}\], is


The general solution of the differential equation \[\frac{dy}{dx} = e^{x + y}\], is


The general solution of a differential equation of the type \[\frac{dx}{dy} + P_1 x = Q_1\] is


\[\frac{dy}{dx} + 1 = e^{x + y}\]


cos (x + y) dy = dx


x2 dy + (x2 − xy + y2) dx = 0


\[\frac{dy}{dx} + 2y = \sin 3x\]


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


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


`2 cos x(dy)/(dx)+4y sin x = sin 2x," given that "y = 0" when "x = pi/3.`


`(dy)/(dx)+ y tan x = x^n cos x, n ne− 1`


For the following differential equation, find the general solution:- \[\frac{dy}{dx} = \sqrt{4 - y^2}, - 2 < y < 2\]


For the following differential equation, find the general solution:- `y log y dx − x dy = 0`


Solve the following differential equation:-

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


Find a particular solution of the following differential equation:- x2 dy + (xy + y2) dx = 0; y = 1 when x = 1


Solve the differential equation : `("x"^2 + 3"xy" + "y"^2)d"x" - "x"^2 d"y" = 0  "given that"  "y" = 0  "when"  "x" = 1`.


Solution of the differential equation `"dx"/x + "dy"/y` = 0 is ______.


The general solution of the differential equation x(1 + y2)dx + y(1 + x2)dy = 0 is (1 + x2)(1 + y2) = k.


Find the general solution of `"dy"/"dx" + "a"y` = emx 


Find the general solution of `(x + 2y^3)  "dy"/"dx"` = y


Solve the differential equation dy = cosx(2 – y cosecx) dx given that y = 2 when x = `pi/2`


tan–1x + tan–1y = c is the general solution of the differential equation ______.


The general solution of `("d"y)/("d"x) = 2x"e"^(x^2 - y)` is ______.


General solution of `("d"y)/("d"x) + ytanx = secx` is ______.


The solution of the differential equation `x(dy)/("d"x) + 2y = x^2` is ______.


General solution of `("d"y)/("d"x) + y` = sinx is ______.


The solution of the differential equation `("d"y)/("d"x) = (x + 2y)/x` is x + y = kx2.


Find a particular solution satisfying the given condition `- cos((dy)/(dx)) = a, (a ∈ R), y` = 1 when `x` = 0


Find the particular solution of the differential equation `x (dy)/(dx) - y = x^2.e^x`, given y(1) = 0.


Find the general solution of the differential equation:

`log((dy)/(dx)) = ax + by`.


The curve passing through (0, 1) and satisfying `sin(dy/dx) = 1/2` is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×