SOLUTION:
Case: Initial value problem
An initial value problem is an ordinary differential equation together with an initial condition that specifies the value of the unknown function at a given point in the domain
To solve the initial value problem, we have a value of the derivative of y when x is known.
y'(x)= value
Given:
f(4)=2
[tex]\begin{gathered} f^{\prime}(x)=\frac{1}{x}-2x+x^{\frac{1}{2}} \\ Integrating \\ f(x)=ln(x)-\frac{2x^2}{2}+\frac{2x^{\frac{3}{2}}}{3}+C \\ f(x)=ln(x)-x^2+\frac{2x^{\frac{3}{2}}}{3}+C \\ f(4)=2 \\ f(4)=ln(4)-2(4)^2+\frac{2(2)^{\frac{3}{2}}}{3}+C \\ 2=1.386-32+1.8856+C \\ C=2-1.3863+32-1.8856 \\ C=30.7281 \end{gathered}[/tex]The resulting equation will be:
[tex]f(x)=\ln x-x^2+\frac{2x^{\frac{3}{2}}}{3}+30.7281[/tex]Final answer:
[tex]f(x)=\operatorname{\ln}x-x^2+\frac{2x^{\frac{3}{2}}}{3}+30.7281[/tex]