Advertisements
Advertisements
Question
A solution of differential equation which can be obtained from the general solution by giving particular values to the arbitrary constant is called ______ solution
Advertisements
Solution
particular
RELATED QUESTIONS
If 1, `omega` and `omega^2` are the cube roots of unity, prove `(a + b omega + c omega^2)/(c + s omega + b omega^2) = omega^2`
Solve the equation for x: `sin^(-1) 5/x + sin^(-1) 12/x = π/2, x ≠ 0`
Verify that y = − x − 1 is a solution of the differential equation (y − x) dy − (y2 − x2) dx = 0.
Show that the differential equation of which \[y = 2\left( x^2 - 1 \right) + c e^{- x^2}\] is a solution is \[\frac{dy}{dx} + 2xy = 4 x^3\]
(sin x + cos x) dy + (cos x − sin x) dx = 0
dy + (x + 1) (y + 1) dx = 0
Solve the following differential equation:
\[y\left( 1 - x^2 \right)\frac{dy}{dx} = x\left( 1 + y^2 \right)\]
(x + y) (dx − dy) = dx + dy
Solve the following initial value problem:
\[x\frac{dy}{dx} + y = x \cos x + \sin x, y\left( \frac{\pi}{2} \right) = 1\]
In a simple circuit of resistance R, self inductance L and voltage E, the current `i` at any time `t` is given by L \[\frac{di}{dt}\]+ R i = E. If E is constant and initially no current passes through the circuit, prove that \[i = \frac{E}{R}\left\{ 1 - e^{- \left( R/L \right)t} \right\}.\]
The differential equation obtained on eliminating A and B from y = A cos ωt + B sin ωt, is
Form the differential equation of the family of circles having centre on y-axis and radius 3 unit.
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.
x2y dx − (x3 + y3) dy = 0
Solve the following differential equation.
`dy/dx + y` = 3
`dy/dx = log x`
Solve the differential equation `("d"y)/("d"x) + y` = e−x
Solve the following differential equation y2dx + (xy + x2) dy = 0
Solve the following differential equation `("d"y)/("d"x)` = x2y + y
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`
Solve `x^2 "dy"/"dx" - xy = 1 + cos(y/x)`, x ≠ 0 and x = 1, y = `pi/2`
lf the straight lines `ax + by + p` = 0 and `x cos alpha + y sin alpha = p` are inclined at an angle π/4 and concurrent with the straight line `x sin alpha - y cos alpha` = 0, then the value of `a^2 + b^2` is
