For each of the following differential equations, a particular solution can be found by inspection. Obtain such a solution [math]y_p[/math], and also find the general solution.
- [math](D^2+3D-10)y=5[/math]
- [math](D^2+1)y = 2x[/math]
- [math]\deriv2y - 4y = 12x - 20[/math]
- [math]\deriv2y + \dydx - 2y = -2x^2+6x-4[/math]
- [math](D^2-2D-3)y=e^x[/math]
- [math]\deriv2y - 2\dydx + y = 8e^{3x}[/math]
- [math]D(D^2-9)y = 2e^{-x}[/math]
- [math](D^2+4)y = 3\sin x[/math]
- [math](D^2+4)y = 3\sin x + 4x + 8[/math]
- [math]\deriv2y + 3\dydx = 5\cos x - 5 \sin x[/math]
- [math](D^2+3)y = 5\cos 3x[/math]
- [math](D^2+2D-2)y = 13 \cos 2x[/math].
Find the particular solution [math]f(x)[/math] of the differential equation [math](D^2+1)y=2x[/math] which has the property that [math]f(0)=3[/math] and [math]f^\prime(0)=2[/math]. (Hint: Find the general solution first and then apply the given boundary conditions to find the values of the constants.)
Find the particular solution [math]y(x)[/math] of the differential equation
with the property that [math]y(0)=\frac12[/math] and [math]y^\prime(0)=\frac12[/math].
The current [math]i[/math] in a given alternating-current circuit is a function of time [math]t[/math] and is governed by the differential equation
Find [math]i[/math] as a function of [math]t[/math], if [math]i=0[/math] and [math]\nxder{}it = 6[/math] when [math]t=0[/math].
Find the general solution of each of the following differential equations.
- [math](D-2)^2y=4x^2-5[/math]
- [math](D^2-3D+2)y=4x+3[/math]
- [math]\deriv2y + \dydx - 2y = 5e^{-x}[/math]
- [math]D(D-2)y = 6x^2+2x-6[/math]
- [math](D^2+D-2)y = 6e^{-2x}[/math]
- [math](D^2+D-2)y=6e^{-2x} + 2x - 4[/math]
- [math](D^2+D-2)y = 6e^{-2x} + 15e^x[/math]
- [math]D^2(D+3)y = 5x-2[/math]
- [math]\deriv2y + 4y = 5 \cos 3x[/math]
- [math]\deriv2y + 9y = 2 \sin 3x[/math]
- [math](D^2+1)y = 10 \sin x + 3e^{-x}[/math]
- [math](D^2+1)y = 4\sin x + 8\cos x[/math]
- [math](D^2-2D+1)y = 3e^x \sin x[/math]
- [math](D^2+2D+2)y = 3e^x \cos x[/math].