Advertisements
Advertisements
प्रश्न
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).
Advertisements
उत्तर
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
संबंधित प्रश्न
Verify that y = cx + 2c2 is a solution of the differential equation
Verify that \[y = ce^{tan^{- 1}} x\] is a solution of the differential equation \[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + \left( 2x - 1 \right)\frac{dy}{dx} = 0\]
Verify that \[y = e^{m \cos^{- 1} x}\] satisfies the differential equation \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} - m^2 y = 0\]
Differential equation \[\frac{d^2 y}{d x^2} + y = 0, y \left( 0 \right) = 0, y' \left( 0 \right) = 1\] Function y = sin x
xy (y + 1) dy = (x2 + 1) dx
(y + xy) dx + (x − xy2) dy = 0
Find the particular solution of the differential equation \[\frac{dy}{dx} = - 4x y^2\] given that y = 1, when x = 0.
x2 dy + y (x + y) dx = 0
The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also find the time necessary for the number of bacteria to be 10 times the number of initial present.
The slope of the tangent at a point P (x, y) on a curve is \[\frac{- x}{y}\]. If the curve passes through the point (3, −4), find the equation of the curve.
Find the equation of the curve which passes through the point (2, 2) and satisfies the differential equation
\[y - x\frac{dy}{dx} = y^2 + \frac{dy}{dx}\]
The integrating factor of the differential equation \[x\frac{dy}{dx} - y = 2 x^2\]
Solve the following differential equation : \[\left( \sqrt{1 + x^2 + y^2 + x^2 y^2} \right) dx + xy \ dy = 0\].
Solve the following differential equation.
`(dθ)/dt = − k (θ − θ_0)`
For each of the following differential equations find the particular solution.
`y (1 + logx)dx/dy - x log x = 0`,
when x=e, y = e2.
Solve the following differential equation.
xdx + 2y dx = 0
Solve the following differential equation.
`dy/dx + y` = 3
Solve the following differential equation.
`dy/dx + 2xy = x`
The solution of `dy/ dx` = 1 is ______.
The solution of `dy/dx + x^2/y^2 = 0` is ______
Choose the correct alternative.
The solution of `x dy/dx = y` log y is
Solve the differential equation `("d"y)/("d"x) + y` = e−x
Solve `("d"y)/("d"x) = (x + y + 1)/(x + y - 1)` when x = `2/3`, y = `1/3`
Solve the differential equation (x2 – yx2)dy + (y2 + xy2)dx = 0
Solve: `("d"y)/("d"x) + 2/xy` = x2
State whether the following statement is True or False:
The integrating factor of the differential equation `("d"y)/("d"x) - y` = x is e–x
Find the particular solution of the following differential equation
`("d"y)/("d"x)` = e2y cos x, when x = `pi/6`, y = 0.
Solution: The given D.E. is `("d"y)/("d"x)` = e2y cos x
∴ `1/"e"^(2y) "d"y` = cos x dx
Integrating, we get
`int square "d"y` = cos x dx
∴ `("e"^(-2y))/(-2)` = sin x + c1
∴ e–2y = – 2sin x – 2c1
∴ `square` = c, where c = – 2c1
This is general solution.
When x = `pi/6`, y = 0, we have
`"e"^0 + 2sin pi/6` = c
∴ c = `square`
∴ particular solution is `square`
An appropriate substitution to solve the differential equation `"dx"/"dy" = (x^2 log(x/y) - x^2)/(xy log(x/y))` is ______.
There are n students in a school. If r % among the students are 12 years or younger, which of the following expressions represents the number of students who are older than 12?
A man is moving away from a tower 41.6 m high at a rate of 2 m/s. If the eye level of the man is 1.6 m above the ground, then the rate at which the angle of elevation of the top of the tower changes, when he is at a distance of 30 m from the foot of the tower, is
`d/(dx)(tan^-1 (sqrt(1 + x^2) - 1)/x)` is equal to:
Which of the following defines a differential equation?
