Question

slove d^3y/dx^3+6d^2y/dx^2+11dy/dx+6y=0​

Answer 1

Answer:

pls see the attachment

Answer 2

The required solution is

$\bf y = c_1 {e}^{ - x} + c_2 {e}^{ - 2x} + c_3 {e}^{ -3 x}$

Where c₁ , c₂ , c₃ are arbitrary constants

Given :

$\displaystyle \sf{ \frac{ {d}^{3}y }{d {x}^{3} } +6\frac{ {d}^{2}y }{d {x}^{2} } + 11 \frac{ {d}^{}y }{d {x}^{} } + 6y = 0 }$

To find :

The solution

Solution :

Step 1 of 3 :

Find roots of the auxiliary equation

Here the given differential equation is

$\displaystyle \sf{ \frac{ {d}^{3}y }{d {x}^{3} } +6\frac{ {d}^{2}y }{d {x}^{2} } + 11 \frac{ {d}^{}y }{d {x}^{} } + 6y = 0 }$

$\sf Let \: y = {e}^{mx} \: be \: the \: trial \: solution$

Then the auxiliary equation is

m³ + 6m² + 11m + 6 = 0

Step 2 of 3 :

Find roots of the auxiliary equation

$\sf {m}^{3} + 6 {m}^{2} + 11m + 6 = 0$

$\displaystyle \sf{ \implies (m +1 )(m + 2)(m + 3) = 0}$

$\displaystyle \sf{ \implies m = - 1, - 2, - 3 }$

Step 3 of 3 :

Find the required solution

Hence the required solution is

$\sf y = c_1 {e}^{ - x} + c_2 {e}^{ - 2x} + c_3 {e}^{ -3 x}$

Where c₁ , c₂ , c₃ are arbitrary constants

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