Solve the initial value problem: 16y''-40y'+25y=0 Where the initial conditions are: y(0)=3,\ y'(0)=-\frac{9}{4}

abreviatsjw 2021-12-18 Answered
Solve the initial value problem:
\(\displaystyle{16}{y}{''}-{40}{y}'+{25}{y}={0}\)
Where the initial conditions are:
\(\displaystyle{y}{\left({0}\right)}={3},\ {y}'{\left({0}\right)}=-{\frac{{{9}}}{{{4}}}}\)

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Expert Answer

jean2098
Answered 2021-12-19 Author has 3506 answers

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

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Bubich13
Answered 2021-12-20 Author has 5377 answers
Now we initial Conditions in equation 1 and 2
\(\displaystyle{y}{\left({0}\right)}={\left({c}_{{1}}{x}_{{0}}+{c}_{{2}}\right)}{e}^{{{\frac{{{5}}}{{{4}}}}{x}_{{0}}}}\)
\(\displaystyle{3}={c}_{{2}}{x}_{{1}}\)
\(\displaystyle{c}_{{2}}={3}\)
\(\displaystyle{y}'{\left({0}\right)}={\frac{{{5}}}{{{4}}}}{\left({c}_{{1}}{x}+{c}_{{2}}\right)}{e}^{{{\frac{{{5}}}{{{4}}}}{x}}}+{c}_{{1}}{e}^{{{\frac{{{5}}}{{{4}}}}{x}}}\)
\(\displaystyle=-{\frac{{{9}}}{{{4}}}}={\frac{{{5}}}{{{4}}}}{\left({c}_{{1}}{x}_{{0}}+{c}_{{2}}\right)}{e}^{{{\frac{{{5}}}{{{4}}}}{x}_{{0}}}}+{c}_{{1}}{e}^{{{\frac{{{5}}}{{{4}}}}{x}_{{0}}}}\)
\(\displaystyle=-{\frac{{{9}}}{{{4}}}}={\frac{{{5}}}{{{4}}}}\times{3}+{4}\)
\(\displaystyle=-{\frac{{{9}}}{{{4}}}}={\frac{{{15}}}{{{4}}}}+{c}_{{1}}\)
\(\displaystyle{c}_{{1}}=-{\frac{{{9}}}{{{4}}}}-{\frac{{{15}}}{{{4}}}}\)
\(\displaystyle{c}_{{1}}={\frac{{-{9}-{15}}}{{{4}}}}\)
\(\displaystyle{c}_{{1}}=-{6}\) Solution of the equation is:
\(\displaystyle{y}={\left(-{6}{x}+{3}\right)}{e}^{{{\frac{{{5}}}{{{4}}}}{x}}}\)
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nick1337
Answered 2021-12-28 Author has 10701 answers
Good answers!
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