`y = C_1e^2x + C_2e^(-2x) + C_3sin(2x) + C_4cos(2x)` Determine whether the function is a solution of the differential equation `y^((4))

Given,

$\displaystyle{y}={C}_{{1}}{{e}}^{{2}}{x}+{C}_{{2}}{{e}}^{{-{2}{x}}}+{C}_{{3}}{\sin{{\left({2}{x}\right)}}}+{C}_{{4}}{\cos{{\left({2}{x}\right)}}}$

let us find

$\displaystyle{y}'={\left({C}_{{1}}{{e}}^{{2}}{x}+{C}_{{2}}{{e}}^{{-{2}{x}}}+{C}_{{3}}{\sin{{\left({2}{x}\right)}}}+{C}_{{4}}{\cos{{\left({2}{x}\right)}}}\right)}'$

$\displaystyle={2}{C}_{{1}}{{e}}^{{2}}{x}+{\left(-{2}\right)}{C}_{{2}}{{e}}^{{-{2}{x}}}+{2}{C}_{{3}}{\cos{{\left({2}{x}\right)}}}-{2}{C}_{{4}}{\sin{{\left({2}{x}\right)}}}$

$\displaystyle{y}''={\left({2}{C}_{{1}}{{e}}^{{2}}{x}+{\left(-{2}\right)}{C}_{{2}}{{e}}^{{-{2}{x}}}+{2}{C}_{{3}}{\cos{{\left({2}{x}\right)}}}-{2}{C}_{{4}}{\sin{{\left({2}{x}\right)}}}\right)}'$

$\displaystyle={4}{C}_{{1}}{{e}}^{{2}}{x}+{\left({4}\right)}{C}_{{2}}{{e}}^{{-{2}{x}}}-{4}{C}_{{3}}{\sin{{\left({2}{x}\right)}}}-{4}{C}_{{4}}{\cos{{\left({2}{x}\right)}}}$

$\displaystyle{y}'''={\left({4}{C}_{{1}}{{e}}^{{2}}{x}+{\left({4}\right)}{C}_{{2}}{{e}}^{{-{2}{x}}}-{4}{C}_{{3}}{\sin{{\left({2}{x}\right)}}}-{4}{C}_{{4}}{\cos{{\left({2}{x}\right)}}}\right)}'$

$\displaystyle={8}{C}_{{1}}{{e}}^{{2}}{x}+{\left(-{8}\right)}{C}_{{2}}{{e}}^{{-{2}{x}}}-{8}{C}_{{3}}{\cos{{\left({2}{x}\right)}}}+{8}{C}_{{4}}{\sin{{\left({2}{x}\right)}}}$

$\displaystyle{y}''''={\left({8}{C}_{{1}}{{e}}^{{2}}{x}+{\left(-{8}\right)}{C}_{{2}}{{e}}^{{-{2}{x}}}-{8}{C}_{{3}}{\cos{{\left({2}{x}\right)}}}+{8}{C}_{{4}}{\sin{{\left({2}{x}\right)}}}\right)}'$

$\displaystyle={\left({16}{C}_{{1}}{{e}}^{{2}}{x}+{\left(-{8}\right)}{\left(-{2}\right)}{C}_{{2}}{{e}}^{{-{2}{x}}}-{8}{\left(-{2}\right)}{C}_{{3}}{\sin{{\left({2}{x}\right)}}}+{8}{\left({2}\right)}{C}_{{4}}{\cos{{\left({2}{x}\right)}}}\right)}$

$\displaystyle={\left({16}{C}_{{1}}{{e}}^{{2}}{x}+{16}{C}_{{2}}{{e}}^{{-{2}{x}}}+{16}{C}_{{3}}{\sin{{\left({2}{x}\right)}}}+{16}{C}_{{4}}{\cos{{\left({2}{x}\right)}}}\right)}$

So lets check whether $\displaystyle{y}''''-{16}{y}={0}$ or not

$\displaystyle{\left({16}{C}_{{1}}{{e}}^{{2}}{x}+{16}{C}_{{2}}{{e}}^{{-{2}{x}}}+{16}{C}_{{3}}{\sin{{\left({2}{x}\right)}}}+{16}{C}_{{4}}{\cos{{\left({2}{x}\right)}}}\right)}-{16}{\left({C}_{{1}}{{e}}^{{2}}{x}+{C}_{{2}}{{e}}^{{-{2}{x}}}+{C}_{{3}}{\sin{{\left({2}{x}\right)}}}+{C}_{{4}}{\cos{{\left({2}{x}\right)}}}\right)}$

= $\displaystyle{\left({16}{C}_{{1}}{{e}}^{{2}}{x}+{16}{C}_{{2}}{{e}}^{{-{2}{x}}}+{16}{C}_{{3}}{\sin{{\left({2}{x}\right)}}}+{16}{C}_{{4}}{\cos{{\left({2}{x}\right)}}}\right)}-{\left({16}{C}_{{1}}{{e}}^{{2}}{x}+{16}{C}_{{2}}{{e}}^{{-{2}{x}}}+{16}{C}_{{3}}{\sin{{\left({2}{x}\right)}}}+{16}{C}_{{4}}{\cos{{\left({2}{x}\right)}}}\right)}$

$\displaystyle={0}$

so,

$\displaystyle{y}''''-{16}{y}={0}$

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Wednesday, December 17, 2014

$\displaystyle{y}={C}_{{1}}{{e}}^{{2}}{x}+{C}_{{2}}{{e}}^{{-{2}{x}}}+{C}_{{3}}{\sin{{\left({2}{x}\right)}}}+{C}_{{4}}{\cos{{\left({2}{x}\right)}}}$ Determine whether the function is a solution of the differential equation $\displaystyle{{y}}^{{{\left({4}\right)}}}-{16}{y}={0}$

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Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Wednesday, December 17, 2014

Determine whether the function is a solution of the differential equation

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Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle{y}={C}_{{1}}{{e}}^{{2}}{x}+{C}_{{2}}{{e}}^{{-{2}{x}}}+{C}_{{3}}{\sin{{\left({2}{x}\right)}}}+{C}_{{4}}{\cos{{\left({2}{x}\right)}}}$ Determine whether the function is a solution of the differential equation $\displaystyle{{y}}^{{{\left({4}\right)}}}-{16}{y}={0}$

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?