abreviatsjw

2021-12-18

Solve the initial value problem:

$16y{}^{\u2033}-40{y}^{\prime}+25y=0$

Where the initial conditions are:

$y\left(0\right)=3,\text{}{y}^{\prime}\left(0\right)=-\frac{9}{4}$

Where the initial conditions are:

jean2098

Beginner2021-12-19Added 38 answers

Initial value problem

root are real then solution of intial value problem

Now differentiate then:

Bubich13

Beginner2021-12-20Added 36 answers

Now we initial Conditions in equation 1 and 2

$y\left(0\right)=({c}_{1}{x}_{0}+{c}_{2}){e}^{\frac{5}{4}{x}_{0}}$

$3={c}_{2}{x}_{1}$

${c}_{2}=3$

$y}^{\prime}\left(0\right)=\frac{5}{4}({c}_{1}x+{c}_{2}){e}^{\frac{5}{4}x}+{c}_{1}{e}^{\frac{5}{4}x$

$=-\frac{9}{4}=\frac{5}{4}({c}_{1}{x}_{0}+{c}_{2}){e}^{\frac{5}{4}{x}_{0}}+{c}_{1}{e}^{\frac{5}{4}{x}_{0}}$

$=-\frac{9}{4}=\frac{5}{4}\times 3+4$

$=-\frac{9}{4}=\frac{15}{4}+{c}_{1}$

$c}_{1}=-\frac{9}{4}-\frac{15}{4$

$c}_{1}=\frac{-9-15}{4$

${c}_{1}=-6$ Solution of the equation is:

$y=(-6x+3){e}^{\frac{5}{4}x}$

nick1337

Expert2021-12-28Added 777 answers

Good answers!

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