# Solve the given initial-value problem. [on hold]

y” + y’ + 5y = 0, y(0) = y'(0) = 0

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We solve this by assuming yy is of the form y=Cekxy=C e^{k x}. Plugging in we get
y″+y′+5y=(k2y)+(ky)+5y=(k2+k+5)y=0. y” + y’ + 5y = (k^2 y) + (k y) + 5y = (k^2 + k + 5)y = 0.
This is solved when k2+k+5=0k^2+k+5 = 0, so plugging in the quadratic equation we have this is solved when k=−1±i√192k=\frac{-1\pm i\sqrt{19}}{2}. Subbing back into yy we get y=C1e−1+i√192x+C2e−1−i√192xy = C_1 e^{\frac{-1+ i\sqrt{19}}{2}x} + C_2 e^{\frac{-1- i\sqrt{19}}{2}x}, two terms since there are two possible kk. This can be further simplified into y=e−x/2(c1cos√192x+c2sin√192x)y=e^{-x/2}(c_1 \cos{\frac{\sqrt{19}}{2}x} + c_2 \sin{\frac{\sqrt{19}}{2} x}). Since y(0)=0y(0)=0, set xx to 00 to find y(0)=e−0/2(c1cos√1920+c2sin√1920)=c1y(0)=e^{-0/2}(c_1 \cos{\frac{\sqrt{19}}{2}\,0} + c_2 \sin{\frac{\sqrt{19}}{2}\,0}) = c_1, and so c1=0c_1 = 0. Taking the derivative at zero gives us y′(0)=c2y'(0)=c_2, so c2=0c_2=0.

Unfortunately this means y(x)=e−x/2(0+0)=0y(x) = e^{-x/2}(0+0)=0, so y(x)=0y(x)=0.

I’m think I might have been confusing concepts, but is this right at all? y″=5y−y′y”=5y-y’. Integrating yields y′=52y2−y+c1y’=\frac{5}{2}y^2-y+c_1. Since y′(0)=0y'(0)=0, Then 0=c10=c_1. Then take another integral. You get: y=56y3−12y2+c2.y=\frac{5}{6}y^3-\frac{1}{2}y^2+c_2. Since y(0)=0y(0)=0, then c2=0c_2=0. You then have the equation y=56y3−12y2y=\frac{5}{6}y^3-\frac{1}{2}y^2.
– Aaron M
Oct 21 at 0:57