Practice problem
Solve the differential equation
y′′+y′+y=0y'' + y' + y = 0y′′+y′+y=0
where y(0)=2y(0) = 2y(0)=2 and y′(0)=1y'(0) = 1y′(0)=1.
y(x)=23 e−x2(23sin(32 x)+3cos(32 x))y(x)=\frac23\ e^{-\frac{x}{2}}\bigg(2\sqrt3\sin\Big(\frac{\sqrt3}{2}\ x\Big)+3\cos\Big(\frac{\sqrt3}{2}\ x\Big)\bigg)y(x)=32 e−2x(23sin(23 x)+3cos(23 x))
y(x)=13 e−x2(3sin(23 x)+3cos(23 x))y(x)=\frac13\ e^{-\frac{x}{2}}\bigg(\sqrt3\sin\Big(\frac{\sqrt2}{3}\ x\Big)+3\cos\Big(\frac{\sqrt2}{3}\ x\Big)\bigg)y(x)=31 e−2x(3sin(32 x)+3cos(32 x))
y(x)=34 e−x3(42sin(34 x)+2cos(34 x))y(x)=\frac34\ e^{-\frac{x}{3}}\bigg(4\sqrt2\sin\Big(\frac{\sqrt3}{4}\ x\Big)+2\cos\Big(\frac{\sqrt3}{4}\ x\Big)\bigg)y(x)=43 e−3x(42sin(43 x)+2cos(43 x))
y(x)=14 ex4(5sin(25 x)+5cos(25 x))y(x)=\frac14\ e^{\frac{x}{4}}\bigg(\sqrt5\sin\Big(\frac{\sqrt2}{5}\ x\Big)+5\cos\Big(\frac{\sqrt2}{5}\ x\Big)\bigg)y(x)=41 e4x(5sin(52 x)+5cos(52 x))
Type a concept, like “slope” or “photosynthesis”.