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let's say if we have the differential equation F prime of X is equal to 3x and also we're given the initial condition that F of 0 is equal to 7 with this information find a particular solution to this differential equation so how can we solve that differential equation well let's begin by finding the antiderivative of both sides of that equation so the antiderivative of F prime of X is equal to the antiderivative of 3x so the integration of F prime of X is just going to be f of X and the antiderivative of 3x is going to be 3x squared over 2 plus C now what we need to do is we need to solve for C anytime you want to find a particular solution of a certain differential equation so what we're going to do is plug in the initial condition F of 0 is 7 so 0 is the x value 7 correlates to the Y value or the function value so we're gonna replace the function with 7 so I'm going to write it like this first F of 0 is equal to 3 times 0 squared over 2 plus C now we know that f of 0 is 7 so I can replace this with 7 so 7 is equal to zero squared times derivative of 0 so 7 is equal to C so now I can be write the function replacing this with C so I could say that f of X is equal to 3 over 2 x squared plus C and this is the answer I'm looking for so that is the particular solution to this differential equation now let's go ahead and work on another example let's say that f prime of X is equal to 6x squared minus five and we're given the initial condition that f of 1 is 4 go ahead and solve that differential equation so let's begin by finding the antiderivative of both sides of the function so on the left side is just going to be f of X on the right side the antiderivative of x squared using the power rule you need to add 1 to the exponent 2 plus 1 is 3 and then divide by that number and the antiderivative of 5 is simply 5x and don't forget the constant C so now let's replace X with 1 so f of 1 is going to be 6 divided by 3 which is 2 times 1 to the 3rd minus 5 times 1 plus C now let's replace F of 1 with 4 because f of 1 is equal to 4 so we have 4 is equal to 2 minus 5 plus C 2 minus 5 is negative 3 and to solve for C we need to add 3 to both sides and 4 plus 3 is 7 so C is equal to 7 so now that we have the value of C we can write the final answer so this is the goal if you could solve for this then you can get the answer based on this expression you just need to replace C with 7 so f of X is going to be 6 divided by 2 which is there's supposed to be a 3 by the way 6 divided by 3 which is 2 X cubed minus 5 X plus 7 so this is the solution to the differential equation now instead of being given the first derivative sometimes we may receive the second derivative of the function let's say if the second derivative is 2x minus 3 this time we need two initial conditions 1 for the first derivative and 1 for the original function so if you want to try it go ahead and determine a function for f of X solve this particular differential equation so let's begin by integrating both sides of that equation so the antiderivative of the second derivative F double prime is the first derivative F prime of X and on the right side the antiderivative of X is going to be x squared divided by two and four negative three it becomes negative three X plus some constant C now we need to use this particular initial condition to solve for C so f prime of 1 that's going to be 2 over 2 which is 1 times x squared so that's 1 squared minus 3 times 1 plus C and F prime of 1 is equal to 2 1 squared is 1 3 times 1 is dream and so this is what we have 1 minus 3 is negative 2 so I need to add 2 to both sides 2 plus 2 is 4 so C is equal to 4 so now we can write the general equation for f prime of X so f prime of X is going to be 2 divided by 2 is 1 so that's just x squared minus 3 X + C is 4 so this is the first answer but we need to go all the way to f of X so now let's integrate both sides of that function so we have the antiderivative of F prime of X and that's going to equal the antiderivative of x squared minus 3x plus 4 so on the left this becomes f of X and on the right we have X cubed divided by 3 minus 3x squared over 2 plus 4 X now instead of using C again we're going to use a different constant of integration so let's choose the next letter D now f of 0 is 3 so we have F of 0 and 0 to the third power is 0 0 squared times D over 2 that's 0 and then 4 times 0 plus Adeem now f of 0 is 3 so Dean is equal to 3 so now we could write the final answer and that is that f of X is equal to one-third X cubed minus three over two x squared plus four X and then replace D with three so this is the solution to the differential equation let's try another example so let's say that the second derivative is x squared minus four and F prime of two is three and F of 1 is negative four go ahead and solve the differential equation so let's start by integrating both sides so on the left the antiderivative of the second derivative it's gonna be the first derivative on the right the antiderivative of x squared is X cube over 3 and 4 negative 4 becomes negative 4x and plus C now we know that f prime of 2 is dream so let's replace X with 2 so we have 2 to the third power minus 3 I mean divided by 3 minus 4 times 2 plus C now F prime of 2 is 3 and 2 to the third power is 8 and 4 times 2 is 8 so now let's add 8 to both sides 3 plus 8 is 11 and so we have 11 is equal to 8 over 3 plus C now to get rid of the fraction I'm gonna multiply everything by 3 and so 11 times 3 is 33 8 over 3 times 3 is just 8 and then C times 3 is 3 C so now let's subtract both sides by 8 33 minus 8 is 25 and then we need to divide by 3 so we have C is 25 over 3 so now we can write the final answer f prime of X is equal to 1/3 X cubed minus 4x plus 25 over 3 so if we plugged in 2 we should get 3 so now let's integrate this function so the integration of f prime is simply F and the antiderivative of X cube is going to be X to the fourth divided by four and four exits x squared over two and then for the constant 25 over three it's gonna be twenty five x over three plus a new constant D so now let's determine F of 1 so 1 to the fourth is just one and three times four is 12 and then 4 divided by 2 is 2 times 1 squared which is 1 and then it's gonna be 25 over 3 plus some constant D now let's replace F of 1 with negative 4 so we have 1 over 12 minus 2 plus 25 over 3 plus D now to get rid of all the fractions I'm gonna multiply everything by 12 I guess I don't need this information anymore so negative 4 times 12 that's a negative 48 12 divided by 12 is 1 2 times 12 is 24 12 divided by 3 is 4 times 25 so that's gonna be a hundred and then plus 12 D now a hundred minus 24 is 76 plus one that's gonna be 77 so we have negative 48 is equal to 77 plus 12 D so negative 48 minus 77 that's negative 125 and then we need to divide both sides by 12 so D is negative 125 over 12 now we can write the final answer and that is that f of X is 1 over 12 X to the fourth minus 2x squared plus 25 x over 3 minus 125 over 12 and so this is the solution of the differential equation it was a little bit longer than the other problems but that's how you could find it so that's it for this video thanks for watching