Advertisements
Advertisements
Question
In the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:-
y = ex + 1 y'' − y' = 0
Advertisements
Solution
We have,
y'' − y' = 0 ............(1)
Now,
y = ex +1
⇒ y'= ex
⇒ y'' = ex
Putting the above values in (1), we get
LHS = ex − ex = 0 = RHS
Thus, y = ex + 1 is the solution of the given differential equation.
APPEARS IN
RELATED QUESTIONS
Prove that:
`int_0^(2a)f(x)dx = int_0^af(x)dx + int_0^af(2a - x)dx`
Hence, the given function is the solution to the given differential equation. \[\frac{c - x}{1 + cx}\] is a solution of the differential equation \[(1+x^2)\frac{dy}{dx}+(1+y^2)=0\].
Differential equation \[x\frac{dy}{dx} = 1, y\left( 1 \right) = 0\]
Function y = log x
Differential equation \[\frac{d^2 y}{d x^2} - 3\frac{dy}{dx} + 2y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 3\] Function y = ex + e2x
x cos2 y dx = y cos2 x dy
tan y \[\frac{dy}{dx}\] = sin (x + y) + sin (x − y)
(y2 + 1) dx − (x2 + 1) dy = 0
Solve the following differential equation:
\[y\left( 1 - x^2 \right)\frac{dy}{dx} = x\left( 1 + y^2 \right)\]
Find the particular solution of the differential equation
(1 – y2) (1 + log x) dx + 2xy dy = 0, given that y = 0 when x = 1.
(x2 − y2) dx − 2xy dy = 0
Solve the following initial value problem:-
\[\left( 1 + y^2 \right) dx + \left( x - e^{- \tan^{- 1} y} \right) dx = 0, y\left( 0 \right) = 0\]
Find the equation to the curve satisfying x (x + 1) \[\frac{dy}{dx} - y\] = x (x + 1) and passing through (1, 0).
Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y\] sin x = 1, is
The equation of the curve whose slope is given by \[\frac{dy}{dx} = \frac{2y}{x}; x > 0, y > 0\] and which passes through the point (1, 1) is
The differential equation \[x\frac{dy}{dx} - y = x^2\], has the general solution
Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y \sin x = 1\], is
Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of x-axis.
Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
Solve the following differential equation.
xdx + 2y dx = 0
The solution of `dy/dx + x^2/y^2 = 0` is ______
Solve the differential equation:
`e^(dy/dx) = x`
Solve the following differential equation
`yx ("d"y)/("d"x)` = x2 + 2y2
Solve the following differential equation y log y = `(log y - x) ("d"y)/("d"x)`
Solve the following differential equation y2dx + (xy + x2) dy = 0
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`
The differential equation of all non horizontal lines in a plane is `("d"^2x)/("d"y^2)` = 0
The value of `dy/dx` if y = |x – 1| + |x – 4| at x = 3 is ______.
