Yup, still doing DE!

Calculus Level 4

$\large \dddot y - 2 \ddot y - 5 \dot y +6y=0$

If the above differential equation has the general solution of the form $y = { a }_{ 1 }{ e }^{ at }+{ a }_{ 2 }{ e }^{ bt }+{ a }_{ 3 }{ e }^{ ct }$ , where $a_1$ , $a_2$ and $a_3$ are real constants, and $a$ , $b$ and $c$ are integers, find $a+b+c$ .

Notations: $\dot y =\dfrac { dy }{ dt }$ , $\ddot y =\dfrac { { d }^{ 2 }y }{ d{ t }^{ 2 } }$ , $\dddot y =\dfrac { { d }^{ 3 }y }{ d{ t }^{ 3 } }$ .

The answer is 2.

1 solution

Vishnu C
Apr 2, 2015

$The\quad given\quad DE\quad is\quad a\quad Homogeneous\quad LDE.\\ \\ The\quad characteristic\quad equation\quad is\\ { \lambda }^{ 3 }-2{ \lambda }^{ 2 }-5\lambda +6=0\quad which\quad has\quad a\quad root\quad for\\ \lambda =1.\quad Now,\quad the\quad equation\quad can\quad be\quad factorised\\ to\quad \\ (\lambda -1)(\lambda +2)(\lambda -3)=0.\\ So,\quad the\quad characteristic\quad roots\quad are\quad 1,\quad -2,\quad 3\quad and\\ the\quad general\quad solution\quad is\quad { a }_{ 1 }{ e }^{ 1t }{ +a }_{ 2 }{ e }^{ -2t }{ +a }_{ 3 }{ e }^{ 3t }.\\ \therefore \quad a+b+c=\boxed { 2 }$

$a+b+ c$ is the sum of the roots which can be directly concluded to be $2$ .

Sudeep Salgia - 6 years, 2 months ago

did not understand.......

rajat kharbanda - 6 years, 1 month ago

Yup, still doing DE!

The answer is 2.

1 solution

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