Advertisements
Advertisements
प्रश्न
Show that the function y = A cos x + B sin x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + y = 0\]
Advertisements
उत्तर
We have,
\[y = A \cos x + B \sin x............(1)\]
Differentiating both sides of equation (1) with respect to x, we get
\[\frac{dy}{dx} = - A \sin x + B \cos x...........(2)\]
Differentiating both sides of equation (2) with respect to x, we get
\[\frac{d^2 y}{d x^2} = - A \cos x - B \sin x\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = - \left( A \cos x + B \sin x \right)\]
\[ \Rightarrow \frac{d^2 y}{d x^2} = - y ...........\left[\text{Using equation }\left( 1 \right) \right]\]
⇒ \[\frac{d^2 y}{d x^2} + y = 0\]
Hence, the given function is the solution to the given differential equation.
APPEARS IN
संबंधित प्रश्न
Verify that y2 = 4a (x + a) is a solution of the differential equations
\[y\left\{ 1 - \left( \frac{dy}{dx} \right)^2 \right\} = 2x\frac{dy}{dx}\]
Verify that \[y = e^{m \cos^{- 1} x}\] satisfies the differential equation \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} - m^2 y = 0\]
(sin x + cos x) dy + (cos x − sin x) dx = 0
Solve the differential equation \[x\frac{dy}{dx} + \cot y = 0\] given that \[y = \frac{\pi}{4}\], when \[x=\sqrt{2}\]
In a bank principal increases at the rate of r% per year. Find the value of r if ₹100 double itself in 10 years (loge 2 = 0.6931).
If y(x) is a solution of the different equation \[\left( \frac{2 + \sin x}{1 + y} \right)\frac{dy}{dx} = - \cos x\] and y(0) = 1, then find the value of y(π/2).
(x + y) (dx − dy) = dx + dy
3x2 dy = (3xy + y2) dx
Solve the following differential equations:
\[\frac{dy}{dx} = \frac{y}{x}\left\{ \log y - \log x + 1 \right\}\]
Solve the following initial value problem:-
\[\frac{dy}{dx} + y\cot x = 2\cos x, y\left( \frac{\pi}{2} \right) = 0\]
The rate of growth of a population is proportional to the number present. If the population of a city doubled in the past 25 years, and the present population is 100000, when will the city have a population of 500000?
The decay rate of radium at any time t is proportional to its mass at that time. Find the time when the mass will be halved of its initial mass.
Find the equation of the curve such that the portion of the x-axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2).
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.
The normal to a given curve at each point (x, y) on the curve passes through the point (3, 0). If the curve contains the point (3, 4), find its equation.
The slope of the tangent at each point of a curve is equal to the sum of the coordinates of the point. Find the curve that passes through the origin.
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.
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 \[\frac{du}{dx} + P\left( x \right) u = Q\left( x \right)\]
Which of the following differential equations has y = C1 ex + C2 e−x as the general solution?
Find the equation of the plane passing through the point (1, -2, 1) and perpendicular to the line joining the points A(3, 2, 1) and B(1, 4, 2).
In each of the following examples, verify that the given function is a solution of the corresponding differential equation.
| Solution | D.E. |
| y = ex | `dy/ dx= y` |
For the following differential equation find the particular solution.
`(x + 1) dy/dx − 1 = 2e^(−y)`,
when y = 0, x = 1
Solve the following differential equation.
`dy /dx +(x-2 y)/ (2x- y)= 0`
The integrating factor of the differential equation `dy/dx - y = x` is e−x.
Solve the differential equation:
`e^(dy/dx) = x`
Solve:
(x + y) dy = a2 dx
y2 dx + (xy + x2)dy = 0
`dy/dx = log x`
Solve: `("d"y)/("d"x) + 2/xy` = x2
Given that `"dy"/"dx"` = yex and x = 0, y = e. Find the value of y when x = 1.
Solve the differential equation `dy/dx + xy = xy^2` and find the particular solution when y = 4, x = 1.
