Differential Systems

Calculus Level 5

2 d x d t + 2 x + d y d t − y = 3 t \large 2\frac {dx}{dt} + 2x + \frac{dy}{dt} - y = 3t

d x d t + x + d y d t + y = 1 \large \frac{dx}{dt} + x + \frac{dy}{dt} + y = 1

x ( 0 ) = 1 x(0) = 1 ; y ( 0 ) = 3 y(0) = 3

Given that x x and y y are both functions of t t , the value of d y d x \frac{dy}{dx} when x = 1 + 3 e − 2 e 3 x= 1 + \frac{3}{e} - \frac{2}{e^3} can be expressed in the form − a e b + c a e b − b e d + c \large -\frac{ae^{b} + c}{ae^{b} - be^{d} + c} where a a , b b , c c , and d d are distinct positive integers. Determine a b c d − ( a + b + c + d ) abcd -(a+b+c+d) .


The answer is 24.

Efren Medallo by Efren Medallo , 26, Philippines

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1 solution

2 d x d t + 2 x + d y d t − y = 3 t . . . . . . . . . ( 1 ) d x d t + x + d y d t + y = 1 . . . . . . . . . . ( 2 ) 2\frac { dx }{ dt } +2x+\frac { dy }{ dt } -y=3t\quad \quad \quad \quad \quad .........(1)\\ \frac { dx }{ dt } +x+\frac { dy }{ dt } +y=1\quad \quad \quad \quad \quad \quad \quad ..........(2)

f r o m e q u a t i o n ( 1 ) a n d ( 2 ) w e k n o w t h a t d y d t + 3 y = 2 − 3 t = > y = 1 + 2 e − 3 t − t = > d y d t = − 6 e − 3 t − 1 . . . . . . . . . ( 3 ) from\quad equation\quad (1)\quad and\quad (2)\quad we\quad know\quad that\quad \\ \frac { dy }{ dt } +3y=2-3t\\ =>\quad y=1+2{ e }^{ -3t }-t\\ =>\quad \frac { dy }{ dt } =\quad -6{ e }^{ -3t }-1\quad \quad \quad \quad .........(3)

f r o m ( 2 ) a n d ( 3 ) d x d t + x = 1 + 4 e − 3 t + t = > x = t − 2 e − 3 t + 3 e − t from\quad (2)\quad and\quad (3)\\ \frac { dx }{ dt } +x=1+4{ e }^{ -3t }+t\\ =>\quad x=t-2{ e }^{ -3t }+3{ e }^{ -t }

w h e n x = 1 + 3 e − 2 e 3 t = 1 = > d x d t = 1 + 6 e − 3 t − 3 e t when\quad x=1+\frac { 3 }{ e } -\frac { 2 }{ { e }^{ 3 } } \quad t=1\\ =>\quad \frac { dx }{ dt } =1+6{ e }^{ -3t }-3{ e }^{ t }

a t t = 1 d y d x = − [ 1 + 6 e − 3 1 + 6 e − 3 − 6 e − 1 ] = − [ e 3 + 6 e 3 − 3 e 2 + 6 ] at\quad t=1\quad \frac { dy }{ dx } =-\left[ \frac { 1+6{ e }^{ -3 } }{ 1+6{ e }^{ -3 }-6{ e }^{ -1 } } \right] =-\left[ \frac { { e }^{ 3 }\quad +\quad 6 }{ { e }^{ 3 }\quad -3{ e }^{ 2 }\quad +\quad 6\quad } \right]

− [ e 3 + 6 e 3 − 3 e 2 + 6 ] = − [ a e b + c a e 3 − b e d + c ] a = 1 b = 3 c = 6 d = 2 a b c d − ( a + b + c + d ) = 36 − 12 = 24 -\left[ \frac { { e }^{ 3 }\quad +\quad 6 }{ { e }^{ 3 }\quad -3{ e }^{ 2 }\quad +\quad 6\quad } \right] =-\left[ \frac { { ae }^{ b }\quad +\quad c }{ { ae }^{ 3 }\quad -b{ e }^{ d }\quad +\quad c\quad } \right] \\ a=1\quad b=\quad 3\quad c=6\quad d=2\\ abcd\quad -(a+b+c+d)=36-12=24

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