Advertisements
Advertisements
प्रश्न
Solve the following differential equation.
`(x + a) dy/dx = – y + a`
उत्तर
`(x + a) dy/dx = – y + a`
∴ `dy/dx + y/((x+a)) = a / ((x+a))`
The given equation is of the form
`dy/ dx + py = Q`
where, `P = 1/((x+a)) and Q = a/((x+a))`
∴ I.F. = `e ^(int^(pdx) = e ^(int^(1/(x+a))^dx)`
= `e^(log^ |x+a|) = (x+a)`
∴ Solution of the given equation is
`y ( I.F.) = int Q (I.F.) dx + c `
∴ `y(x + a) = int a/((x+a)) (x+a) dx + c`
∴ `y(x + a) = a int 1 dx + c`
∴ y (x + a) = ax + c
APPEARS IN
संबंधित प्रश्न
Form the differential equation representing the family of ellipses having centre at the origin and foci on x-axis.
Show that the differential equation of which \[y = 2\left( x^2 - 1 \right) + c e^{- x^2}\] is a solution is \[\frac{dy}{dx} + 2xy = 4 x^3\]
C' (x) = 2 + 0.15 x ; C(0) = 100
tan y \[\frac{dy}{dx}\] = sin (x + y) + sin (x − y)
dy + (x + 1) (y + 1) dx = 0
Solve the differential equation \[\left( 1 + x^2 \right)\frac{dy}{dx} + \left( 1 + y^2 \right) = 0\], given that y = 1, when x = 0.
y ex/y dx = (xex/y + y) dy
Solve the following initial value problem:-
\[\frac{dy}{dx} + y\cot x = 2\cos x, y\left( \frac{\pi}{2} \right) = 0\]
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 \[i = \frac{E}{R}\left\{ 1 - e^{- \left( R/L \right)t} \right\}.\]
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 \[\frac{du}{dx} + P\left( x \right) u = Q\left( x \right)\]
Determine the order and degree of the following differential equations.
Solution | D.E. |
y = 1 − logx | `x^2(d^2y)/dx^2 = 1` |
Solve the following differential equation.
(x2 − y2 ) dx + 2xy dy = 0
State whether the following is True or False:
The integrating factor of the differential equation `dy/dx - y = x` is e-x
The function y = ex is solution ______ of differential equation
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