Advertisements
Advertisements
Question
In a bank principal increases at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0.5 = 1.648).
Solution
Let at any instant t, the principal be P .
Here, it is given that the principal increases at the rate of 5 % per year .
\[\frac{dP}{dt} = \frac{5P}{100}\]
\[ \Rightarrow \frac{dP}{P} = \frac{1}{20}dt\]
Integrating both sides, we get
\[\ln P = \frac{t}{20} + \ln C ...........(1) \]
Initially at t = 0, it is given that P = Rs 1000 .
\[\ln 1000 = \ln C\]
Substituting the value of ln C in (1), we get
\[\ln P = \frac{t}{20} + \ln 1000\]
\[\text{ Putting }t = 10, \text{ we get }\]
\[\ln \frac{P}{1000} = 0 . 5\]
\[ \Rightarrow \frac{P}{1000} = e^{0 . 5} \]
\[ \Rightarrow P = 1000 \times 1 . 648\]
\[ = 1648\]
Therefore, Rs 1000 will be worth Rs 1648 after 10 years .
APPEARS IN
RELATED QUESTIONS
Show that the differential equation of which y = 2(x2 − 1) + \[c e^{- x^2}\] is a solution, is \[\frac{dy}{dx} + 2xy = 4 x^3\]
Verify that \[y = ce^{tan^{- 1}} x\] is a solution of the differential equation \[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + \left( 2x - 1 \right)\frac{dy}{dx} = 0\]
For the following differential equation verify that the accompanying function is a solution:
Differential equation | Function |
\[x + y\frac{dy}{dx} = 0\]
|
\[y = \pm \sqrt{a^2 - x^2}\]
|
tan y dx + sec2 y tan x dy = 0
Find the solution of the differential equation cos y dy + cos x sin y dx = 0 given that y = \[\frac{\pi}{2}\], when x = \[\frac{\pi}{2}\]
Solve the following initial value problem:-
\[\frac{dy}{dx} + 2y = e^{- 2x} \sin x, y\left( 0 \right) = 0\]
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 \[x = \frac{\pi}{2}\]
Experiments show that radium disintegrates at a rate proportional to the amount of radium present at the moment. Its half-life is 1590 years. What percentage will disappear in one year?
The slope of a curve at each of its points is equal to the square of the abscissa of the point. Find the particular curve through the point (−1, 1).
Find the equation of the curve that passes through the point (0, a) and is such that at any point (x, y) on it, the product of its slope and the ordinate is equal to the abscissa.
Write the differential equation obtained by eliminating the arbitrary constant C in the equation x2 − y2 = C2.
Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y\] sin x = 1, is
The integrating factor of the differential equation \[x\frac{dy}{dx} - y = 2 x^2\]
Show that y = ae2x + be−x is a solution of the differential equation \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} - 2y = 0\]
Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of x-axis.
Solve the following differential equation.
`dy/dx + y` = 3
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.
Solve:
(x + y) dy = a2 dx
Select and write the correct alternative from the given option for the question
The differential equation of y = Ae5x + Be–5x is
For the differential equation, find the particular solution
`("d"y)/("d"x)` = (4x +y + 1), when y = 1, x = 0
Solve the following differential equation y2dx + (xy + x2) dy = 0
Verify y = log x + c is the solution of differential equation `x ("d"^2y)/("d"x^2) + ("d"y)/("d"x)` = 0
Solve the following differential equation
`y log y ("d"x)/("d"y) + x` = log y
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`
Solve the following differential equation
sec2 x tan y dx + sec2 y tan x dy = 0
Solution: sec2 x tan y dx + sec2 y tan x dy = 0
∴ `(sec^2x)/tanx "d"x + square` = 0
Integrating, we get
`square + int (sec^2y)/tany "d"y` = log c
Each of these integral is of the type
`int ("f'"(x))/("f"(x)) "d"x` = log |f(x)| + log c
∴ the general solution is
`square + log |tan y|` = log c
∴ log |tan x . tan y| = log c
`square`
This is the general solution.