Advertisements
Advertisements
Question
The differential equation of `y = k_1e^x+ k_2 e^-x` is ______.
Options
`(d^2y)/dx^2 - y = 0`
`(d^2y)/dx^2 + dy/dx = 0`
`(d^2y)/dx^2 + ydy/dx = 0`
`(d^2y)/dx^2 + y = 0`
Advertisements
Solution
The differential equation of `y = k_1e^x+ k_2 e^-x` is `underlinebb((d^2y)/dx^2 - y = 0)`.
Explanation:
`y = k_1e^x+ k_2 e^-x`
Differentiating w.r.t. x, we get
`dy/dx = k_1e^x - k_2 e^-x`
Again, differentiating w.r.t. x, we get
`(d^2y)/dx^2 = k_1 e^x + k_2 e^-x`
∴ `(d^2y)/dx^2 = y`
∴ `(d^2y)/dx^2 - y = 0`
RELATED QUESTIONS
Show that the differential equation of which y = 2(x2 − 1) + \[c e^{- x^2}\] is a solution, is \[\frac{dy}{dx} + 2xy = 4 x^3\]
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x\frac{dy}{dx} = y\]
|
y = ax |
(ey + 1) cos x dx + ey sin x dy = 0
(y + xy) dx + (x − xy2) dy = 0
A curve is such that the length of the perpendicular from the origin on the tangent at any point P of the curve is equal to the abscissa of P. Prove that the differential equation of the curve is \[y^2 - 2xy\frac{dy}{dx} - x^2 = 0\], and hence find the curve.
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.
Show that all curves for which the slope at any point (x, y) on it is \[\frac{x^2 + y^2}{2xy}\] are rectangular hyperbola.
Write the differential equation obtained eliminating the arbitrary constant C in the equation xy = C2.
Which of the following transformations reduce the differential equation \[\frac{dz}{dx} + \frac{z}{x}\log z = \frac{z}{x^2} \left( \log z \right)^2\] into the form \[\frac{du}{dx} + P\left( x \right) u = Q\left( x \right)\]
Solve the following differential equation : \[y^2 dx + \left( x^2 - xy + y^2 \right)dy = 0\] .
Solve the following differential equation.
(x2 − y2 ) dx + 2xy dy = 0
Solve the following differential equation.
`dy/dx + y` = 3
Solve the following differential equation.
`dy/dx + 2xy = x`
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 the differential equation:
`e^(dy/dx) = x`
Solve:
(x + y) dy = a2 dx
x2y dx – (x3 + y3) dy = 0
`xy dy/dx = x^2 + 2y^2`
`dy/dx = log x`
The function y = ex is solution ______ of differential equation
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
The differential equation (1 + y2)x dx – (1 + x2)y dy = 0 represents a family of:
Solve the differential equation `dy/dx + xy = xy^2` and find the particular solution when y = 4, x = 1.
