Question

solve the linear differential equations d^3y/dx^3+6(d^2y/dx^2)+11(dy/dx)+6y=0.

Answer 1

Assume,

$\longrightarrow y=e^{kx}$

Then,

$\longrightarrow y_1=ke^{kx}$

and,

$\longrightarrow y_2=k^2e^{kx}$

and,

$\longrightarrow y_3=k^3e^{kx}$

So,

$\longrightarrow y_3+6y_2+11y_1+6y=0$

$\longrightarrow k^3e^{kx}+6k^2e^{kx}+11ke^{kx}+6e^{kx}=0$

$\longrightarrow e^{kx}\left(k^3+6k^2+11k+6\right)=0$

Since $e^{kx}\neq0,$

$\longrightarrow k^3+6k^2+11k+6=0$

$\longrightarrow k^3+k^2+5k^2+5k+6k+6 =0$

$\longrightarrow k^2(k+1)+5k(k+1)+6(k+1)=0$

$\longrightarrow (k+1)(k^2+5k+6)=0$

$\longrightarrow (k+1)(k^2+2k+3k+6)=0$

$\longrightarrow (k+1)(k(k+2)+3(k+2))=0$

$\longrightarrow (k+1)(k+2)(k+3)=0$

$\Longrightarrow k\in\{-3,\ -2,\ -1\}$

Hence the possible solutions are,

$\longrightarrow y=e^{-3x}$

$\longrightarrow y=e^{-2x}$

$\longrightarrow y=e^{-x}$

Generally, we can represent these solutions as,

$\longrightarrow\underline{\underline{y=Ae^{-3x}+Be^{-2x}+Ce^{-x}}}$

for some arbitrary constants $A,\ B$ and $C.$

Answer 2

${\colorbox{cyan}{✿QuesTion}}$

solve the linear differential equations :-

$\frac{d^3y}{dx^3} +6( \frac{d^2y/dx^2}{?} )+11 \frac{dy}{dx} +6y=0.$

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${\colorbox{pink}{✿ AnsWer}}$

$( {D}^{3} y \: + 6 {D}^{2} y \: + 11Dy + 6y) = 0$

$❆( \frac{d}{dx} = D)$

$\purple \therefore$

$\longmapsto \: ( {D}^{3} + 6 {D}^{2} + 11D + 6)y = 0$

$\phi(D)y = 0$

$\red {A.E.}$

$: \longrightarrow \phi \: (D) = 0$

$\blue{i.e.}$

$: \longrightarrow {D}^{3} + 6 {D}^{2} + 11D + 6 = 0$

$\orange{roots:-}$

$\longmapsto \: D \: = - 1, \: \: D \: = - 2, \: D \: = - 3.$

$\pink{G.S. \: is :-}$

${ \boxed{ \underline{ \orange{y \: = c_1 \: {e}^{ - x} + c _2 \: {e}^{ - 2x} + c_ 3 \: {e}^{ - 3x} }}}}$

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