Question

Solve the initial-value problem

1. y" - 3y' - 10y = 0, y(0) = 0, y'(0) = 7.

2. y"+14y' + 50y = 0,y(0) = 2, y'(0) = -17

3. 6y"-y'-y = 0, y(0) = 10, y'(0) = 0

4. 6y"+y'-y = 0, y(0) = -1, y'(0) = 3

Answer 1

1.

$k^2-3k-10=0$

$k=\frac{3\pm \sqrt{9+40}}{2}$

$k_1=-2,k_2=5$

$y=c_1e^{-2x}+c_2e^{5x}$

$0=c_1+c_2$

$y'=-2c_1e^{-2x}+5c_2e^{5x}$

$7=-2c_1+5c_2$

$2c_2+5c_2=7$

$c_2=1,c_1=-1$

$y=-e^{-2x}+e^{5x}$

2.

$k^2+14k+50=0$

$k=\frac{-14\pm \sqrt{196-200}}{2}$

$k_{1,2}=\frac{-14\pm 2i}{2}=-7\pm i$

$y=e^{-7x}(c_1cosx+c_2sinx)$

$2=c_1$

$y'=-7e^{-7x}(c_1cosx+c_2sinx)+e^{-7x}(c_2cosx-c_1sinx)$

$-17=-7c_1+c_2$

$c_2=-3$

$y=e^{-7x}(2cosx-3sinx)$

3.

$6k^2-k-1=0$

$k=\frac{1\pm \sqrt{1+24}}{12}$

$k_1=-\frac{1}{3},k_2=\frac{1}{2}$

$y=c_1e^{-x/3}+c_2e^{x/2}$

$10=c_1+c_2$

$y'=-\frac{c_1e^{-x/3}}{3}+\frac{c_2e^{x/2}}{2}$

$0=-c_1/3+c_2/2$

$\frac{c_2-10}{3}+\frac{c_2}{2}=0$

$c_2=4,c_1=6$

$y=6e^{-x/3}+4e^{x/2}$

4.

$6k^2+k-1=0$

$k=\frac{-1\pm \sqrt{1+24}}{12}$

$k_1=-\frac{1}{2},k_2=\frac{1}{3}$

$y=c_1e^{-x/2}+c_2e^{x/3}$

$-1=c_1+c_2$

$y'=-\frac{c_1e^{-x/2}}{2}+\frac{c_2e^{x/3}}{3}$

$3=-\frac{c_1}{2}+\frac{c_2}{3}$

$3(c_2+1)+2c_2=18$

$c_2=3,c_1=-4$

$y=-4e^{-x/2}+3e^{x/3}$