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
Show that the differential equation of which y = 2(x2 − 1) + \[c e^{- x^2}\] is a solution, is \[\frac{dy}{dx} + 2xy = 4 x^3\]
Show that y = AeBx is a solution of the differential equation
Verify that y = − x − 1 is a solution of the differential equation (y − x) dy − (y2 − x2) dx = 0.
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x + y\frac{dy}{dx} = 0\]
|
\[y = \pm \sqrt{a^2 - x^2}\]
|
xy (y + 1) dy = (x2 + 1) dx
(1 + x) (1 + y2) dx + (1 + y) (1 + x2) dy = 0
Solve the differential equation \[\frac{dy}{dx} = \frac{2x\left( \log x + 1 \right)}{\sin y + y \cos y}\], given that y = 0, when x = 1.
If the marginal cost of manufacturing a certain item is given by C' (x) = \[\frac{dC}{dx}\] = 2 + 0.15 x. Find the total cost function C (x), given that C (0) = 100.
Find the curve for which the intercept cut-off by a tangent on x-axis is equal to four times the ordinate of the point of contact.
Show that the equation of the curve whose slope at any point is equal to y + 2x and which passes through the origin is y + 2 (x + 1) = 2e2x.
At every point on a curve the slope is the sum of the abscissa and the product of the ordinate and the abscissa, and the curve passes through (0, 1). Find the equation of the curve.
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.
Find the equation of the curve passing through the point (0, 1) if the slope of the tangent to the curve at each of its point is equal to the sum of the abscissa and the product of the abscissa and the ordinate of the point.
If sin x is an integrating factor of the differential equation \[\frac{dy}{dx} + Py = Q\], then write the value of P.
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
Which of the following is the integrating factor of (x log x) \[\frac{dy}{dx} + y\] = 2 log x?
Verify that the function y = e−3x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + \frac{dy}{dx} - 6y = 0.\]
Find the particular solution of the differential equation `"dy"/"dx" = "xy"/("x"^2+"y"^2),`given that y = 1 when x = 0
In the following example, verify that the given function is a solution of the corresponding differential equation.
| Solution | D.E. |
| y = xn | `x^2(d^2y)/dx^2 - n xx (xdy)/dx + ny =0` |
Solve the following differential equation.
`dy /dx +(x-2 y)/ (2x- y)= 0`
Solve the following differential equation.
y dx + (x - y2 ) dy = 0
The solution of `dy/ dx` = 1 is ______
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 the differential equation:
dr = a r dθ − θ dr
Solve
`dy/dx + 2/ x y = x^2`
Select and write the correct alternative from the given option for the question
The differential equation of y = Ae5x + Be–5x is
Solve the following differential equation `("d"y)/("d"x)` = x2y + y
