4a^3+b^2+2a^2b^2+3ab/ab

1 answer:

8 0

\left[a _{3}\right] = \left[ \frac{ - b^{2}}{6}+\frac{\frac{ - b^{4}}{3}+\left( \frac{-1}{3}\,i \right) \,\sqrt{3}\,b^{4}}{2^{\frac{2}{3}}\,\sqrt[3]{\left( -1296 - 432\,b^{2} - 16\,b^{6}+\sqrt{\left( 1679616+1119744\,b^{2}+186624\,b^{4}+41472\,b^{6}+13824\,b^{8}\right) }\right) }}+\frac{\frac{ - \sqrt[3]{\left( -1296 - 432\,b^{2} - 16\,b^{6}+\sqrt{\left( 1679616+1119744\,b^{2}+186624\,b^{4}+41472\,b^{6}+13824\,b^{8}\right) }\right) }}{24}+\left( \frac{1}{24}\,i \right) \,\sqrt{3}\,\sqrt[3]{\left( -1296 - 432\,b^{2} - 16\,b^{6}+\sqrt{\left( 1679616+1119744\,b^{2}+186624\,b^{4}+41472\,b^{6}+13824\,b^{8}\right) }\right) }}{\sqrt[3]{2}}\right][a3]=⎣⎢⎢⎢⎢⎡6−b2+2323√(−1296−432b2−16b6+√(1679616+1119744b2+186624b4+41472b6+13824b8))3−b4+(3−1i)√3b4+3√224−3√(−1296−432b2−16b6+√(1679616+1119744b2+186624b4+41472b6+13824b8))+(241i)√33√(−1296−432b2−16b6+√(1679616+1119744b2+186624b4+41472b6+13824b8))⎦⎥⎥⎥⎥⎤

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3. Solve the differential equations a) y'' + 12y' + 32y = 0 b) y'' + 14y' + 49y = 0 c) y'' + 10y' + 34y = 0, y(0) = 1, y'(0) = 4

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$y(x)=3e^{-4x}-2e^{-8x}$

Step-by-step explanation:

I will do the first one thoroughly so you won't have any problems following to complete the rest of them.

This is a linear homogeneous second order differential, so to solve it we will use:

$y(x)=C_{1}e^{r_{1}x}+C_{2}e^{r_{2}x}$ which is a theorem that says that if r1x and r2x are both solutions off a linear homogeneous equation, and C1 and C2 are any constants, then the function above is also a solution of the equation.