# Given f(x) = 2x , f(2) = 1 , f(1) = 3 , find the particular solution to the differential equation.

We are given f''(x)=2x, f'(2)=1, and f(1)=3 and we are asked to find the particular solution to the differential equation.

Since we know the second derivative, we can use the indefinite integral (or antidifferentiation) to find that `f'(x)=x^2+C` . C is called the constant of integration; we need it because a family of functions has the same derivative where the family of functions differ only by a constant.

Since we are given f'(2)=1, we can solve for C. Substituting 2 for x and 1 for y we get 1=4+C ==> C=-3. So `f'(x)=x^2-3` .

We can integrate again to find f(x). `f(x)=1/3x^3-3x+C`

Since we know f(1)=3 we can again substitute to find the value of this constant of integration. 3=1/3(1)^3-3(1)+C.

3=-8/3+C ==> C=17/3

So, the original equation, and particular solution to the given differential equation, is

`f(x)=1/3x^3-3x+17/3`

Check:

f(1)=1/3(1)^3-3(1)+17/3=1/3-9/3+17/3=9/3=3 as required.

The first derivative is f'(x)=x^2-3

f'(2)=2^2-3=4-3=1 as required.

f''(x)=2x as required.

Note that we could have found the general solution by integrating (taking antiderivatives.)

f''(x)=2x

Use the general power rule `int u^ndu=1/(n+1)u^(n+1)+C`

`f'(x)=x^2+C_1`

`f''(x)=1/3x^3+C_1x+C_2`

To see why we need the "C", note that the first derivative of x^2,x^2+1,x^2-17, etc., all are 2x. The C is a vertical translation of the graph of the function and thus does not change the first derivative. They will have the same critical points and the same behavior over the same intervals.

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