Question

Guess a solution to y'' + y' + y = 5.

Answer 1

Answer:

general solution=$e^{-\frac{1}{2} x}(Acos\frac{3}{2} x+Bsin\frac{3}{2}x)$+5

Step-by-step explanation:

using linear differential equation method

y'' + y' + y = 5

writing down the characteristics equation.

$m^2+m+1=0$

using quadratic formula

$m=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

we get

$m=\frac{-1\pm \sqrt{1^2-4(1)(1)}}{2(1)}$

$m=-\frac{1}{2} \pm \frac{3}{2} i$

now Complementary function(CF)

$y=e^{ax}(Acosbx+Bsinbx)\\y=e^{-\frac{1}{2} x}(Acos\frac{3}{2} x+Bsin\frac{3}{2}x)$

now for particular integrals

$D^2y+Dy+y=5\\(D^2+D+1)y=5\\y=\frac{5}{D^2+D+1}$

$P.I.=\frac{5\times e^{0x} }{D^2+D+1}$

putting D=0

we get

P.I.=5

general solution=CF+PI

general solution=$e^{-\frac{1}{2} x}(Acos\frac{3}{2} x+Bsin\frac{3}{2}x)$+5