Looks like either [tex]f''(x)=4x^2[/tex] or [tex]f''(x)=\frac4{x^2}[/tex]...
In the first case, integrate both sides twice to get
[tex]f''(x)=4x^2\implies f'(x)=\dfrac43x^3+C_1\implies f(x)=\dfrac13x^4+C_1x+C_2[/tex]
Then the initial conditions give
[tex]f'(1)=2\implies 2=\dfrac43\cdot1^3+C_1\implies C_1=\dfrac23[/tex]
[tex]f(1)=5\implies 5=\dfrac13\cdot1^4+C_1\cdot1+C_2\implies C_2=4[/tex]
so that the particular solution is
[tex]f(x)=\dfrac{x^4}3+\dfrac{2x}3+4[/tex]
If instead [tex]f''(x)=\frac4{x^2}[/tex], we have
[tex]f''(x)=\dfrac4{x^2}\implies f'(x)=-\dfrac4x+C_1\implies f(x)=-4\ln|x|+C_1x+C_2[/tex]
[tex]f'(1)=2\implies 2=-\dfrac41+C_1\implies C_1=6[/tex]
[tex]f(1)=5\implies 5=-4\ln|1|+C_1\cdot1+C_2\implies C_2=-1[/tex]
[tex]\implies f(x)=-4\ln|x|+6x-1[/tex]
