Find the particular solution of the differential equation f''(x) = 2 that satisfies the initial conditions f'(2) = 5 and f(2) = 10.

Step-by-step solution

1. 1

   Plan the two antiderivatives and the two constants. You are given the second derivative, so recovering $f$ takes two integrations, and each one introduces one arbitrary constant. That is exactly why the problem supplies two conditions — one to kill each constant. Integrate in the order $f''\to f'\to f$ and pin down each constant as soon as it appears.

2. 2

   Integrate once to get $f'$. The antiderivative of the constant $2$ is $2x$:

   $f'(x)=\int 2\,dx=2x+C_{1}$

3. 3

   Use $f'(2)=5$ to find $C_{1}$. Substituting $x=2$:

   $f'(2)=2(2)+C_{1}=4+C_{1}$

   Setting that equal to $5$ gives $C_{1}=1$, so

   $f'(x)=2x+1$

   Fixing $C_1$ now, rather than carrying it along, keeps the second integration clean.

4. 4

   Integrate again to get $f$. Antidifferentiate $2x+1$ term by term:

   $f(x)=\int (2x+1)\,dx=x^{2}+x+C_{2}$

5. 5

   Use $f(2)=10$ to find $C_{2}$. Substituting $x=2$:

   $f(2)=2^{2}+2+C_{2}=6+C_{2}$

   Setting that equal to $10$ gives $C_{2}=4$, so the particular solution is

   $f(x)=x^{2}+x+4$

6. 6

   Verify all three requirements. Differentiating gives $f'(x)=2x+1$ and $f''(x)=2$, so the differential equation holds. At $x=2$: $f'(2)=2(2)+1=5$ and $f(2)=4+2+4=10$. Both initial conditions check out exactly.