# Given the function y(x)C_1e^x+C_2e^x \cos xa) Prove that it is

Given the function $y\left(x\right){C}_{1}{e}^{x}+{C}_{2}{e}^{x}\mathrm{cos}x$
a) Prove that it is a solution to the equation $y{}^{‴}-3y{}^{″}+4{y}^{\prime }-2y=0$
b) Show that the initial value problem has no solution.

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accimaroyalde
$y\left(x\right){C}_{1}{e}^{x}+{C}_{2}{e}^{x}\mathrm{cos}x$
${y}^{\prime }={C}_{1}{e}^{x}+{C}_{2}\left[-{e}^{x}\mathrm{sin}x+{e}^{x}\mathrm{cos}x\right]$
$y{}^{″}={C}_{1}{e}^{x}+{C}_{2}\left[-{e}^{x}\mathrm{cos}x-{e}^{x}\mathrm{sin}x-{e}^{x}\mathrm{sin}x+{e}^{x}\mathrm{cos}x\right]$
$y{}^{‴}={C}_{1}{e}^{x}+{C}_{2}\left[{e}^{x}\mathrm{sin}x-{e}^{x}\mathrm{cos}x-{e}^{x}\mathrm{cos}x-{e}^{x}\mathrm{sin}x-{e}^{x}\mathrm{sin}x-{e}^{x}\mathrm{cos}x-{e}^{x}\mathrm{sin}x+{e}^{x}\mathrm{cos}x\right]$
$\therefore {y}^{\prime }={C}_{1}{e}^{x}+{C}_{2}\left[{e}^{x}\mathrm{cos}x-{e}^{x}\mathrm{sin}x\right]$
$y{}^{″}{C}_{1}{e}^{x}+{C}_{2}\left[-2{e}^{x}\mathrm{sin}x\right]$
$y{}^{‴}={C}_{1}{e}^{x}+{C}_{2}\left[-2{e}^{x}\mathrm{cos}x-2{e}^{x}\mathrm{sin}x\right]$
Substitute in $y{}^{″}-3y{}^{″}+4{y}^{\prime }-2y$
$⇒{C}_{1}{e}^{x}-2{C}_{2}{e}^{x}\mathrm{cos}x-2{C}_{2}{e}^{x}\mathrm{sin}x-3{C}_{1}{e}^{x}+6{C}_{2}{e}^{x}\mathrm{sin}x+4{C}_{1}{e}^{x}+4{C}_{2}{e}^{x}\mathrm{cos}x-4{C}_{2}{e}^{x}\mathrm{sin}x-2{C}_{1}{e}^{x}-2{C}_{2}{e}^{x}\mathrm{cos}x$
$=0$
Since all terms cancel out. Hence given y is solution.
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Terry Ray
$y\left(x\right)={C}_{1}{e}^{x}+{C}_{2}{e}^{x}\mathrm{cos}x$
$y\left(0\right)=1⇒{C}_{1}+{C}_{2}=1$
${y}^{\prime }\left(0\right)=0⇒{C}_{1}=0$
$y{}^{″}\left(0\right)=0⇒{C}_{1}+{C}_{2}\left[-2\right]=0$
$⇒{c}_{1}=2{C}_{2}$, since ${C}_{1}=0⇒{C}_{2}=0$
But ${C}_{1}+{C}_{2}\ne 1$
Hence no solution is possible.