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X
- WE WANT TO SOLVE THE SEPARABLE DIFFERENTIAL EQUATION DXDT
= 9 DIVIDED BY X
AND TO FIND THE PARTICULAR SOLUTION
SATISFYING THE INITIAL CONDITIONS X OF 0 = 7.
SO BECAUSE WE CAN SOLVE THIS DIFFERENTIAL EQUATION
USING SEPARATION OF VARIABLES.
ON ONE SIDE OF THE EQUATION WE WANT A FUNCTION OF X x DX.
ON THE OTHER SIDE OF THE EQUATION
WE WANT A FUNCTION OF T x DT.
SO STARTING WITH DXDT = 9 DIVIDED BY X.
LET'S WRITE THIS IN DIFFERENTIAL FORM
OR MULTIPLY BOTH SIDES BY DT.
THAT WOULD GIVE US DX = 9 DIVIDED BY X x DT.
NOW BECAUSE THE DX IS ON THE LEFT SIDE
WE WANT THE VARIABLE X ON THE LEFT SIDE.
SO LET'S GO AHEAD AND MULTIPLY BOTH SIDES OF THE EQUATION BY X
GIVING US XDX = 9 DT.
NOW THAT WE HAVE THE EQUATION IN THIS FORM
WE CAN INTEGRATE BOTH SIDES OF THE EQUATION.
THE INTEGRAL OF X TO THE FIRST WITH RESPECT TO X
WOULD BE X SQUARED DIVIDED BY 2 + A CONSTANT OF INTEGRATION
BUT WE'LL GO AHEAD AND INCLUDE THE CONSTANT
WITH THE CONSTANT ON THE RIGHT.
SO WE HAVE = THE INTEGRAL OF 9 WITH RESPECT TO T WOULD BE 9T
+ A CONSTANT OF INTEGRATION WHICH WE'LL CALL C SUB 1
WHICH AGAIN INCLUDES THE CONSTANT FROM THE LEFT
AND THE RIGHT.
NOW FOR THE NEXT STEP
LET'S GO AHEAD AND SOLVE THIS FOR X SQUARED.
SO WE'LL MULTIPLY BOTH SIDES OF THE EQUATION BY 2.
SO WE'D HAVE X SQUARED = 18T.
THEN WE'D HAVE + 2 x C SUB 1.
BUT 2 x C SUB 1 IS JUST ANOTHER CONSTANT.
SO LET'S LET C = 2 x C SUB 1.
SO WE CAN JUST WRITE + C.
AND NOW TO SOLVE FOR X
WE WOULD SQUARE ROOT BOTH SIDES OF THE EQUATION.
AND IN MOST CASES WE'D HAVE A + OR - HERE ON THE RIGHT.
SO WE WOULD HAVE X = + OR - THE SQUARE ROOT OF 18T + C.
BUT NOTICE HOW WE'RE GIVEN THAT X OF 0 = 7.
SO WHEN T IS 0 THE X VALUE IS ONLY +7 AND NOT -7
AND THEREFORE WE DON'T NEED THIS + OR -.
WE WOULD ONLY HAVE THE POSITIVE SQUARE ROOT
OR THE PRINCIPAL SQUARE ROOT.
SO IN OUR CASE WE HAVE X = THE POSITIVE SQUARE ROOT OF 18T + C.
SO NOT ONLY DOES X OF 0 = 7
TELL US WE'RE GOING TO HAVE A PRINCIPAL SQUARE ROOT HERE,
IT'S ALSO GOING TO ALLOW US TO FIND THE VALUE OF C,
THIS CONSTANT HERE, WHICH WILL GIVE US THE PARTICULAR SOLUTION.
SO IF WE KNOW THE GENERAL SOLUTION
IS X = THE SQUARE ROOT OF 18T + C.
AND IF X OF 0 = 7
THAT MEANS THAT THIS FUNCTION WOULD CONTAIN THE POINT
WITH A (T) COORDINATE OF (0) AND A (X) COORDINATE OF (7).
SO THIS ALL FOR C WILL SUBSTITUTE 0 FOR T AND 7 FOR X.
SO WE'D HAVE 7 EQUALS--AND NOW IF T IS 0, 18, 2 WOULD BE 0
SO WE WOULD HAVE JUST A SQUARE ROOT OF C.
AND NOW WE'LL SQUARE BOTH SIDES OF THE EQUATION TO SOLVE FOR C.
SO WE HAVE 49 = C.
WHICH MEANS THE PARTICULAR SOLUTION
TO THIS GIVEN DIFFERENTIAL EQUATION
WITH THE INITIAL CONDITIONS X OF 0 = 7
IS X OF T = THE SQUARE ROOT OF 18T + 49.
I HOPE YOU FOUND THIS HELPFUL.