Prove that these two triangles are congruent.

Mathematics

1 answer:

fredd [130]3 years ago

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Answer:

thank for the pointshhhhfgtf5ttctctx4dd5y8hhousjshs

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Y" - 4y = (x2 - 3) sin 2x

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$y''-4y=0$

has characteristic equation

$r^2-4=0$

which has roots at $r=\pm2$ , giving the characteristic solution

$y_c=C_1e^{2x}+C_2e^{-2x}$

For the nonhomogeneous part of the ODE, let $y_p=(a_2x^2+a_1x+a_0)\sin2x+(b_2x^2+b_1x+b_0)\cos2x$ . Then

${y_p}''=(-4b_2x^2+(8a_2-b_1)x+4a_1-4b_0+2b_2)\cos2x+(-4a_2x^2+(-4a_1-8b_2)x-4a_0+2a_2-4b_1)\sin2x$

Substituting into the ODE gives

$(-8b_2x^2+(8a_2-b_1)x+4a_1-8b_0+2b_2)\cos2x+(-8a_2x^2+(-8a_1-8b_2)x-8a_0+2a_2-4b_1)\sin2x=(x^2-3)\sin2x$

It follows that

$\begin{cases}-8b_2=0\\8a_2-8b_1=0\\4a_1-8b_0+2b_2=0\\-8a_2=1\\-8a_1-8b_2=0\\-8a_0+2a_2-4b_1=-3\end{cases}\implies\begin{cases}a_2=-\dfrac18\\\\a_1=0\\\\a_0=\dfrac{13}{32}\\\\b_2=0\\\\b_1=-\dfrac18\\\\b_0=0\end{cases}$

which yields the particular solution

$y_p=-\dfrac18x^2\sin2x+\dfrac{13}{32}\sin2x-\dfrac18x\cos2x$

So the general solution is

$y=y_c+y_p$
$y=C_1e^{2x}+C_2e^{-2x}-\dfrac18x^2\sin2x+\dfrac{13}{32}\sin2x-\dfrac18x\cos2x$

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