Advertisements
Advertisements
प्रश्न
C' (x) = 2 + 0.15 x ; C(0) = 100
उत्तर
\[C' \left( x \right) = 2 + 0 . 15x\]
\[ \Rightarrow \frac{dC}{dx} = 2 + 0 . 15x\]
\[ \Rightarrow dC = \left( 2 + 0 . 15x \right)dx\]
Integrating both sides, we get
\[\int dC = \int\left( 2 + 0 . 15x \right) dx\]
\[ \Rightarrow C = 2x + \frac{0 . 15}{2} x^2 + D . . . . . \left( 1 \right)\]
\[\text{ It is given that C }\left( 0 \right) = 100 . \]
\[ \therefore 100 = 2\left( 0 \right) + \frac{0 . 15}{2}\left( 0 \right) + D\]
\[ \Rightarrow D = 100\]
\[\text{ Substituting the value of D in } \left( 1 \right), \text{ we get }\]
\[C = 2x + \frac{0 . 15}{2} x^2 + 100\]
\[\text{ Hence, }C = 2x + \frac{0 . 15}{2} x^2 + 100 \text{ is the solution to the given differential equation .}\]
APPEARS IN
संबंधित प्रश्न
Show that the function y = A cos 2x − B sin 2x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + 4y = 0\].
Verify that y2 = 4ax is a solution of the differential equation y = x \[\frac{dy}{dx} + a\frac{dx}{dy}\]
Show that y = e−x + ax + b is solution of the differential equation\[e^x \frac{d^2 y}{d x^2} = 1\]
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[y = \left( \frac{dy}{dx} \right)^2\]
|
\[y = \frac{1}{4} \left( x \pm a \right)^2\]
|
Differential equation \[\frac{dy}{dx} + y = 2, y \left( 0 \right) = 3\] Function y = e−x + 2
Differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 2\] Function y = xex + ex
(sin x + cos x) dy + (cos x − sin x) dx = 0
(ey + 1) cos x dx + ey sin x dy = 0
(1 + x) (1 + y2) dx + (1 + y) (1 + x2) dy = 0
Solve the following differential equation:
\[\left( 1 + y^2 \right) \tan^{- 1} xdx + 2y\left( 1 + x^2 \right)dy = 0\]
Find the particular solution of the differential equation \[\frac{dy}{dx} = \frac{xy}{x^2 + y^2}\] given that y = 1 when x = 0.
Solve the following initial value problem:
\[x\frac{dy}{dx} + y = x \cos x + \sin x, y\left( \frac{\pi}{2} \right) = 1\]
Solve the following initial value problem:-
\[\frac{dy}{dx} + y\cot x = 2\cos x, y\left( \frac{\pi}{2} \right) = 0\]
If the interest is compounded continuously at 6% per annum, how much worth Rs 1000 will be after 10 years? How long will it take to double Rs 1000?
The rate of increase of bacteria in a culture is proportional to the number of bacteria present and it is found that the number doubles in 6 hours. Prove that the bacteria becomes 8 times at the end of 18 hours.
Which of the following is the integrating factor of (x log x) \[\frac{dy}{dx} + y\] = 2 log x?
What is integrating factor of \[\frac{dy}{dx}\] + y sec x = tan x?
Determine the order and degree of the following differential equations.
| Solution | D.E. |
| y = 1 − logx | `x^2(d^2y)/dx^2 = 1` |
For the following differential equation find the particular solution.
`dy/ dx = (4x + y + 1),
when y = 1, x = 0
The differential equation of `y = k_1e^x+ k_2 e^-x` is ______.
Choose the correct alternative.
The integrating factor of `dy/dx - y = e^x `is ex, then its solution is
Select and write the correct alternative from the given option for the question
Bacterial increases at the rate proportional to the number present. If original number M doubles in 3 hours, then number of bacteria will be 4M in
Choose the correct alternative:
General solution of `y - x ("d"y)/("d"x)` = 0 is
Verify y = log x + c is the solution of differential equation `x ("d"^2y)/("d"x^2) + ("d"y)/("d"x)` = 0
Solve `x^2 "dy"/"dx" - xy = 1 + cos(y/x)`, x ≠ 0 and x = 1, y = `pi/2`
Integrating factor of the differential equation `"dy"/"dx" - y` = cos x is ex.
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
Solve the differential equation
`y (dy)/(dx) + x` = 0
Solve the differential equation `dy/dx + xy = xy^2` and find the particular solution when y = 4, x = 1.
