# Find the general solutions of the differential equations 6y^{4} +11y"+4y

Find the general solutions of the differential equations $6{y}^{4}+11y4y=0$
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Mary Nicholson
$6{y}^{4}+11y4y=0$ (1)
Eq (1) can be written as
$\left(6{D}^{4}+11{D}^{2}+4\right)y=0$
AE. $6{m}^{4}+11{m}^{2}+4=0$
For value of m we put ${m}^{2}=u$
${m}^{4}={4}^{2}$
$6{u}^{2}+11u+4=0$

Which is comple roots.
$y={c}_{1}\mathrm{cos}\left(\sqrt{\frac{1}{2}t}\right)+{c}_{2}\mathrm{sin}\left(\sqrt{\frac{1}{2}}t\right)+{c}_{3}\mathrm{cos}\left(\frac{2t}{\sqrt{3}}\right)+{c}_{4}\mathrm{sin}\left(\frac{2t}{\sqrt{3}}\right)$
###### Not exactly what you’re looking for?
Mary Nicholson
The auxilary equation is,
$6{m}^{4}+11{m}^{2}+4=0$
$⇒{m}^{2}=\frac{-11±\sqrt{121-96}}{12}=\frac{-1}{2},\frac{-4}{3}$
$⇒m=±\frac{i}{\sqrt{2}},±\frac{2i}{\sqrt{3}}$
If $\alpha ±i\beta$ are the roots of auxilary equation, then the solution is $y={e}^{\alpha x}\left({c}_{1}\mathrm{cos}\beta x+{c}_{2}\mathrm{sin}\beta x\right)$
Thus, the required solution is,
$y=\left({c}_{1}\mathrm{cos}\frac{x}{\sqrt{2}}+{c}_{2}\mathrm{sin}\frac{x}{\sqrt{2}}\right)+\left({c}_{3}\mathrm{cos}\frac{2x}{\sqrt{3}}+{c}_{4}\mathrm{sin}\frac{2x}{\sqrt{3}}\right)$
###### Not exactly what you’re looking for?
karton

$\begin{array}{}m=\frac{-6±\sqrt{{6}^{2}-4\left(1\right)\left(11\right)}}{2\left(1\right)}\\ =\frac{-6±\sqrt{-8}}{2}\\ =\frac{-6±i\sqrt{8}}{2}\\ =\frac{-6±2i\sqrt{2}}{2}\\ =-3±i\sqrt{2}\\ y={e}^{-3x}\left({C}_{1}\mathrm{sin}\sqrt{2x}+{C}_{2}\mathrm{cos}\sqrt{2x}\right)\end{array}$