abreviatsjw

2021-12-18

Solve the initial value problem:
$16y{}^{″}-40{y}^{\prime }+25y=0$
Where the initial conditions are:

jean2098

Initial value problem
$16y{}^{″}-40{y}^{\prime }+25y=0$ We can find solution of the equation:
$16{D}^{2}-40D+25=0$ where $D=\frac{d}{dx}$
$16{D}^{2}-20D-20D+25=0$
$=40\left(40-5\right)-5\left(40-5\right)=0$
$=\left(40-5\right)\left(40-5\right)=0$
$=D=\frac{5}{4};\frac{5}{4}$
root are real then solution of intial value problem
$y=\left({c}_{1}x+{c}_{2}\right){e}^{\frac{5}{4}x}$...(1)
Now differentiate then:
${y}^{\prime }=\frac{5}{4}\left({c}_{1}x+{c}_{2}\right){e}^{\frac{5}{4}x}+{c}_{1}{e}^{\frac{5}{4}}$...(2)

Bubich13

Now we initial Conditions in equation 1 and 2
$y\left(0\right)=\left({c}_{1}{x}_{0}+{c}_{2}\right){e}^{\frac{5}{4}{x}_{0}}$
$3={c}_{2}{x}_{1}$
${c}_{2}=3$
${y}^{\prime }\left(0\right)=\frac{5}{4}\left({c}_{1}x+{c}_{2}\right){e}^{\frac{5}{4}x}+{c}_{1}{e}^{\frac{5}{4}x}$
$=-\frac{9}{4}=\frac{5}{4}\left({c}_{1}{x}_{0}+{c}_{2}\right){e}^{\frac{5}{4}{x}_{0}}+{c}_{1}{e}^{\frac{5}{4}{x}_{0}}$
$=-\frac{9}{4}=\frac{5}{4}×3+4$
$=-\frac{9}{4}=\frac{15}{4}+{c}_{1}$
${c}_{1}=-\frac{9}{4}-\frac{15}{4}$
${c}_{1}=\frac{-9-15}{4}$
${c}_{1}=-6$Solution of the equation is:
$y=\left(-6x+3\right){e}^{\frac{5}{4}x}$

nick1337