- WELCOME TO AN INTRODUCTION TO DIFFERENTIAL EQUATIONS.
THE GOALS OF THIS VIDEO ARE TO DETERMINE
A GENERAL AND A PARTICULAR SOLUTION
TO A DIFFERENTIAL EQUATION IN THE FORM OF DYDX = F OF X.
A DIFFERENTIAL EQUATION IS AN EQUATION
THAT CONTAINS DERIVATIVES AND HERE ARE THREE EXAMPLES.
TO SOLVE A DIFFERENTIAL EQUATION WE WANT TO DETERMINE AN EQUATION
THAT SATISFIES THE DERIVATIVE EQUATION
AND THIS WILL REQUIRE INTEGRATION.
SO TO SOLVE A DIFFERENTIAL EQUATION IN THIS FORM,
IT'S HELPFUL TO WRITE IT IN LEIBNIZ'S NOTATION
AND THEN IN DIFFERENTIAL FORM.
REMEMBER IF DYDX = F OF X, THEN DY = F OF X DX.
ONCE IT'S IN THIS FORM
WE CAN INTEGRATE BOTH SIDES OF THE EQUATION,
WHICH WILL ALLOW US TO SOLVE THE DIFFERENTIAL EQUATION.
LET'S GO AHEAD AND TAKE A LOOK AT HOW THIS IS DONE.
SO THE GOAL HERE IS TO FIND A FUNCTION
THAT SATISFIES THIS DIFFERENTIAL EQUATION.
EVEN THOUGH YOU MAY BE ABLE TO SOLVE THIS JUST BY LOOKING,
LETS GO AHEAD SHOW THE STEPS.
THE FIRST THING WE DO IS REWRITE THIS IN LEIBNIZ'S NOTATION.
SO WE'D HAVE DYDX = 2X AND THEN IN DIFFERENTIAL FORM
WE WOULD HAVE DY = 2X DX.
NOW WE'LL INTEGRATE BOTH SIDES OF THE EQUATION
AND IT MAY HELP TO PUT A ONE HERE
SO IT CAN THINK THAT WE WANT THE ANTI-DERIVATIVE OF ONE
WITH RESPECTS TO Y.
WELL, THAT WOULD JUST BE Y AND THEN ON THE RIGHT SIDE
THE ANTI-DERIVATIVE OF 2X WITH RESPECTS TO X
WOULD BE 2 x X SQUARED/2 AND THEN + C.
NOW, I COULD HAVE HAD A + C ON THE LEFT AND THE RIGHT
AND THEN COMBINE THEM ON THE RIGHT SIDE,
BUT IT STILL WOULD JUST BE A CONSTANT.
SO LET'S GO AHEAD AND CLEAN THIS UP.
WE HAVE THE FUNCTION Y = X SQUARED + C.
NOTICE OUR SOLUTION IS A FAMILY OF FUNCTIONS
AND THIS IS CALLED A GENERAL SOLUTION
TO A DIFFERENTIAL EQUATION.
SOMETIMES WE'RE GIVEN ENOUGH INFORMATION
TO DETERMINE THE CONSTANT OF INTEGRATION
AS WE SEE HERE THAT Y1 = 3.
SO WE CAN USE THIS INFORMATION TO FIND THE VALUE OF C.
WHEN X = 1, Y = 3.
SO WE WOULD HAVE 3 = 1 SQUARED + C.
SO WE'D HAVE C = 2, WHICH MEANS Y = X SQUARED + 2.
AND THIS IS CALLED THE PARTICULAR SOLUTION.
LET'S GO AHEAD AND TAKE A LOOK AT THE GRAPH
OF WHAT WE JUST FOUND.
ONE WAY TO REPRESENT A DIFFERENTIAL EQUATION
IS BY A SLOPE FIELD AS WE SEE HERE IN RED.
EACH SEGMENT HERE REPRESENTS
THE SLOPE OF THE SOLUTION TO THE DIFFERENTIAL EQUATION.
REMEMBER WE SAID ON THE PREVIOUS SCREEN THAT THE GENERAL SOLUTION
IS Y = X SQUARED + C.
WELL, THIS REPRESENTS A FAMILY OF FUNCTION
FOR DIFFERENT VALUES OF C,
MEANING ANY OF THE PARABOLAS
THAT FIT NICELY IN THIS SLOPE FIELD AS WE SEE HERE,
WOULD BE PART OF THE GENERAL SOLUTION.
BUT SINCE WE WERE GIVEN THAT THE SOLUTION
PASSED THROUGH THE POINT (1, 3) AS WE SEE HERE,
THIS PARABOLA REPRESENTS THE PARTICULAR SOLUTION
Y = X SQUARED + 2.
LET'S GO AHEAD AND TAKE A LOOK AT ANOTHER EXAMPLE.
SO HERE WE'RE GOING TO HAVE DYDX = 1/X,
WHICH MEANS THAT DIFFERENTIAL Y = 1/X x DIFFERENTIAL X.
SO NOW WE CAN INTEGRATE BOTH SIDES OF THE EQUATION.
THE ANTI-DERIVATIVE OF ONE WITH RESPECTS TO Y WOULD BE Y.
THE ANTI-DERIVATIVE OF 1/X WOULD BE NATURAL LOG X
AND THEN OF COURSE WE HAVE THE + C.
AGAIN, THIS IS OUR GENERAL SOLUTION
TO THIS DIFFERENTIAL EQUATION,
BUT NOW WE ARE GIVEN INFORMATION
TO DETERMINE THE PARTICULAR SOLUTION. F OF E = 2.
SO WHEN X = E, Y = 2 AND THAT'LL ALLOW US TO FIND THE VALUE OF C.
SO WE'LL HAVE 2 EQUALS NATURAL LOG E + C.
WELL, NATURAL LOG E IS EQUAL TO ONE.
SO WE HAVE C EQUALS 1,
WHICH TELLS US THAT OUR PARTICULAR SOLUTION
WOULD BE Y EQUALS NATURAL LOG X + 1.
AND AGAIN, LETS TAKE A LOOK AT THE GRAPH OF THAT.
HERE WE HAVE THE SLOPE FIELD FOR THE DIFFERENTIAL EQUATION
AND IN BLACK WE HAVE THE PARTICULAR SOLUTION
BECAUSE WE WERE GIVEN THAT THE SOLUTION WE WANTED
PASSED THROUGH THAT POINT,
THERE'S ONLY ONE FUNCTION THAT FITS NICELY IN THE SLOPE FIELD
AND PASSES THROUGH THAT POINT,
WHICH WE SAID ON THE PREVIOUS SCREEN
WAS Y EQUALS NATURAL LOG X + ONE.
OKAY, SO THE ORDER OF A DIFFERENTIAL EQUATION
IS THE LARGES DERIVATIVE PRESENT IN THE DIFFERENTIAL EQUATION.
THE PREVIOUS EXAMPLES WERE EXAMPLES OF FIRST ORDER
DIFFERENTIAL EQUATIONS.
NOW WE ARE GOING TO TAKE A LOOK
AT A SECOND ORDER DIFFERENTIAL EQUATION.
HERE WE'RE GIVEN F DOUBLE PRIME OF X
IS EQUAL TO X TO THE POWER OF -1/2, F PRIME OF 4 = 2
AND F OF 0 = 1.
SO OUR GOAL IS TO FIND THE ORIGINAL FUNCTION
GIVEN THE SECOND DERIVATIVE.
SO WE PROBABLY CAN MAKE THE CONNECTION NOW
THAT IN ORDER TO FIND THE FIRST DERIVATIVE
WE COULD INTEGRATE THE SECOND DERIVATIVE,
WHICH WOULD BE X TO THE POWER OF -1/2 WITH RESPECTS TO X.
SO IF WE INTEGRATE THE SECOND DERIVATIVE AS WE SEE HERE,
WE'LL FIND THE FIRST DERIVATIVE.
SO THIS WILL BE EQUAL TO ADD ONE TO THE EXPONENT,
THAT'S X TO THE 1/2 DIVIDED BY 1/2 + C,
WHICH IS GOING TO BE EQUAL TO 2X TO THE POWER OF 1/2 + C.
AND AGAIN, THIS IS OUR FIRST DERIVATIVE
SINCE WE STARTED WITH THE SECOND DERIVATIVE.
WELL, NOW WE'RE GIVEN THAT F PRIME OF FOUR IS EQUAL TO TWO.
SO WHEN X = 4, Y = -2 AND THAT WILL ALLOW US TO DETERMINE C.
SO WHEN X = 4, Y = 2.
SO WE WOULD HAVE 2 = 2 x 4 TO THE POWER OF 1/2 + C.
WELL, 4 TO THE 1/2, THAT WOULD BE 2. 2 x 2 = 4.
SO IT LOOKS C = -2.
SO THE FIRST DERIVATIVE FUNCTION IS EQUAL TO 2X
TO THE POWER OF 1/2 - 2.
NOW IF WE INTEGRATE THE FIRST DERIVATIVE,
WE CAN DETERMINE THE ORIGINAL FUNCTION.
LET'S GO AHEAD AND DO THAT ON THE NEXT SCREEN.
SO HERE'S OUR FIRST DERIVATIVE.
SO TO FIND THE ORIGINAL FUNCTION
WE NEED TO INTEGRATE 2X TO THE POWER OF 1/2 - 2
WITH RESPECTS TO X.
SO HERE WE'D HAVE 2 x X TO THE 1/2 + 1,
THAT WOULD BE 3/2 DIVIDED BY 3/2 - 2X + C.
LET'S GO AHEAD AND CLEAN THIS UP.
WE HAVE THE FUNCTION IS EQUAL TO--
THIS WOULD BE 2 x 2/3 = 4/3 X TO THE 3/2 - 2X + C.
BUT WE WERE GIVEN THE INITIAL CONDITION THAT F OF 0 = 1.
SO IF F OF 0 = 1 WE'D HAVE Y = 1 WHEN X = 0.
WELL, THESE TWO TERMS ARE ZERO SO WE HAVE C EQUALS ONE.
SO THE ORIGINAL FUNCTION WOULD BE 4/3X TO THE 3/2 - 2X + 1.
OKAY, THAT'S GOING TO DO IT FOR THIS FIRST VIDEO.
I HOPE IT WAS HELPFUL.