# 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)

###### Not exactly what youâ€™re looking for?
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:
$$\displaystyle{y}={\left(-{6}{x}+{3}\right)}{e}^{{{\frac{{{5}}}{{{4}}}}{x}}}$$
nick1337
Answered 2021-12-28 Author has 10701 answers
Good answers!

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