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X
- WE WANT TO SOLVE THE DIFFERENTIAL EQUATION DYDT = 4
x THE QUANTITY Y - 6.
AND WE'RE ALSO GIVEN THE INITIAL CONDITIONS Y OF 0 = 2.
SO WE'LL BE ABLE TO FIND THE PARTICULAR SOLUTION
TO THIS DIFFERENTIAL EQUATION.
WE CAN SOLVE THIS DIFFERENTIAL EQUATION
USING SEPARATION OF VARIABLES.
MEANING, WE CAN WRITE THIS EQUATION
SO WE HAVE A FUNCTION OF Y x DY ON THE LEFT
AND A FUNCTION OF T x DT ON THE RIGHT.
TO BEGIN LET'S MULTIPLY BOTH SIDES OF THE EQUATION BY DT.
SO WE WOULD HAVE DY = 4 x THE QUANTITY Y - 6 x DT.
AND BECAUSE WE HAVE DY ON THE LEFT
WE WANT THIS FACTOR OF Y - 6 ON THE LEFT.
SO NOW WE'LL DIVIDE BOTH SIDES BY THE QUANTITY Y - 6
THAT WOULD GIVE US DY DIVIDED BY THE QUANTITY Y - 6 = 4 x DT.
IF WE WANT WE CAN WRITE THIS
AS 1 DIVIDED BY THE QUANTITY Y - 6DY = 4DT.
AND NOW IN THIS FORM
WE CAN INTEGRATE BOTH SIDES OF THE EQUATION.
WE'LL INTEGRATE THE LEFT SIDE WITH RESPECT TO Y.
WE'LL INTEGRATE THE RIGHT SIDE WITH RESPECT TO T.
SO THE INTEGRAL OF 1 DIVIDED BY THE QUANTITY Y - 6
IS EQUAL TO NATURAL LOG OF THE QUANTITY Y - 6 + C.
SO I'LL WRITE NATURAL LOG.
I'LL INCLUDE THE ABSOLUTE VALUE HERE
BUT WE'LL SEE IN A LITTLE BIT WHILE IT'S NOT NECESSARY.
AND THEN INSTEAD OF PUTTING PLUS C HERE,
WE'LL COMBINE THE TWO CONSTANTS OF INTEGRATION
AND PUT THE CONSTANT ONLY ON THE RIGHT.
SO THIS IS GOING TO BE EQUAL TO THE INTEGRAL OF 4
WITH RESPECT TO T WOULD BE 4T.
AND THEN WE'LL SAY + C SUB 1,
AGAIN, WHERE C SUB 1 COMBINES THE CONSTANT ON THE LEFT
AND ON THE RIGHT.
NOW OUR GOAL HERE IS TO SOLVE THIS FOR Y
NOT NATURAL LOG OF THE QUANTITY Y - 6.
SO I THINK IN THE PREVIOUS VIDEO WE WROTE THIS LOG EQUATION
AS AN EXPONENTIAL EQUATION
BUT WE CAN ALSO EXPONENTIATE BOTH SIDES OF THE EQUATION.
MEANING E RAISED TO THIS QUANTITY ON THE LEFT
MUST EQUAL E TO THIS QUANTITY ON THE RIGHT.
AND NOTICE HOW E RAISED TO THIS POWER HERE
IS ALWAYS GOING TO BE POSITIVE,
AND THAT'S THE REASON WHY THE QUANTITY Y - 6
WILL ALWAYS BEN POSITIVE.
SO ON THE LEFT THIS SIMPLIFIES NICELY TO JUST Y - 6.
AND ON THE RIGHT WE HAVE E RAISED TO THE POWER OF 4T
+ C SUB 1.
BECAUSE WE HAVE A SUM IN THE EXPONENT
WE CAN RIGHT THIS AS E TO THE POWER OF 4T x E
TO THE POWER OF C SUB 1.
WELL, E TO THE POWER OF C SUB 1 WOULD JUST BE ANOTHER CONSTANT.
SO IF WE LET "A" = E TO THE POWER OF C SUB 1
WE COULD WRITE THIS AS Y - 6 = E TO THE POWER OF 4T x A
OR "A" E TO THE POWER OF 4T.
NOW WE'LL GO AHEAD AND ADD 6 TO BOTH SIDES OF THE EQUATION.
SO WE WOULD HAVE Y = A x E TO THE POWER OF 4T + 6.
SO THIS WOULD BE THE GENERAL SOLUTION
TO THE DIFFERENTIAL EQUATION.
BUT NOW BECAUSE WE HAVE THE INITIAL CONDITIONS Y OF 0 = 2
WE CAN ACTUALLY FIND THE PARTICULAR SOLUTION.
IF Y OF 0 = 2 THAT MEANS WHEN T = 0, Y MUST = 2.
SO BY SUBSTITUTING 2 FOR Y AND 0 FOR T
WE CAN FIND "A" GIVING US THE PARTICULAR SOLUTION.
SO IF WE SUBSTITUTE 2 FOR Y WE'D HAVE 2 = A x--
IF T IS 0 THIS WOULD BE E TO THE 0 WHICH IS 1 + 6.
SUBTRACTING 6 ON BOTH SIDES
NOTICE HOW WE WOULD HAVE "A" = -4
WHICH MEANS THE PARTICULAR SOLUTION
TO THIS DIFFERENTIAL EQUATION
WOULD BE THE FUNCTION Y OF T = "A"
WHICH IS -4 x E RAISE TO THE POWER OF 4T + 6.
I HOPE YOU FOUND THIS HELPFUL.