Question

For the differential equation y" + 4y' + 13y = 0, a general solution is of the form y = e^-2x(C₁sin 3x + C₂cos 3x), where C₁ and C₂ are arbitrary constants. 

Applying the initial conditions y(0) = 4 and y'(0) = 2, find the specific solution. 

y = _______ 

Answer 1

$y''+4y'+13y=0$

find roots first

$x^2+4x+13=0$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$a=1,\:b=4,\:c=13$

$x_{1,\:2}=\frac{-4\pm \sqrt{4^2-4\cdot \:1\cdot \:13}}{2\cdot \:1}$

$\:x_{1,\:2}=\frac{-4\pm \:\sqrt{16-52}}{2}$

$\:x_{1,\:2}=\frac{-4\pm \:\sqrt{-36}}{2}$

$\:x_{1,\:2}=\frac{-4\pm \:6i}{2}$

$\:x_{1,\:2}=-2\pm \:3i$

for complex roots general solution is

$\:y=e^{at}\left(c_1\cos \left(bt\right)+c_2\sin \left(bt\right)\right)$

$y=e^{-2t}\left(c_1\cos \left(3t\right)+c_2\sin \left(3t\right)\right)$ ....................(1)

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here y(0)=4

$4=e^{-2\cdot \:0}\left(c_1\cos \left(3\cdot \:0\right)+c_2\sin \left(3\cdot \:0\right)\right)$

$e^{-2\cdot \:0}\left(c_1\cos \left(3\cdot \:0\right)+c_2\sin \left(3\cdot \:0\right)\right)=4$

$1\cdot \left(\cos \left(3\cdot \:0\right)c_1+\sin \left(3\cdot \:0\right)c_2\right)=4$

$1\cdot \left(\cos \left(0\right)c_1+\sin \left(0\right)c_2\right)=4$

$\left(1c_1+0c_2\right)=4$

$\left(c_1+0\right)=4$

${\color{Red} c_1=4}$

put it back in equation 1

$y=e^{-2t}\left(4\cos \left(3t\right)+c_2\sin \left(3t\right)\right)$ .................(2)

take derivative

$y'= \frac{d}{dt}\left(e^{-2t}\left(4\cos \left(3t\right)+c_2\sin \left(3t\right)\right)\right)$

$y'=\frac{d}{dt}\left(e^{-2t}\right)\left(4\cos \left(3t\right)+c_2\sin \left(3t\right)\right)+\frac{d}{dt}\left(4\cos \left(3t\right)+c_2\sin \left(3t\right)\right)e^{-2t}$

$y'=\left(-2e^{-2t}\right)\left(4\cos \left(3t\right)+c_2\sin \left(3t\right)\right)+\left(-12\sin \left(3t\right)+c_2\cos \left(3t\right)\cdot \:3\right)e^{-2t}$

$y'=-2e^{-2t}\left(4\cos \left(3t\right)+c_2\sin \left(3t\right)\right)+e^{-2t}\left(-12\sin \left(3t\right)+3c_2\cos \left(3t\right)\right)$

here y'(0)=-2

$-2= \:-2e^{-2\cdot 0}\left(4\cos \left(3\cdot 0\right)+c_2\sin \left(3\cdot 0\right)\right)+e^{-2\cdot 0}\left(-12\sin \left(3\cdot 0\right)+3c_2\cos \left(3\cdot 0\right)\right)$

$-2= \:-2e^0\left(4\cos \left(0\right)+c_2\sin \left(0\right)\right)+e^0\left(-12\sin \left(0\right)+3c_2\cos \left(0\right)\right)$

$-2= \:-2\cdot 1\left(4\cdot 1+c_2\cdot 0\right)+1\left(-12\cdot 0+3c_2\cdot 1\right)$

$-2= \:-2\left(4+0\right)+0+3c_2$

$-2=\:-2\left(4\right)+3c_2$

$-2=-8+3c_2$

$-2+8=3c_2$

$6=3c_2$

${\color{Red} c_2=2}$

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put this value in equation 2

$y=e^{-2t}\left(4\cos \left(3t\right)+c_2\sin \left(3t\right)\right)$

${\color{Red} y=e^{-2t}\left(4\cos \left(3t\right)+2\sin \left(3t\right)\right)}$ ..............answer