The differential equation f''(x) = sin(x) with initial conditions f(0) = 0 and f'(0) = 1 is solved correctly as f(x) = -sin(x) + 2x. The solution process involves integrating the second derivative to find the first derivative, applying the initial conditions to determine constants, and integrating again to find the original function. The constants C1 and C2 are calculated as 2 and 0, respectively, confirming the solution is accurate.
PREREQUISITESStudents studying calculus, mathematicians, and engineers who need to solve differential equations and apply initial conditions in their work.
Minor point, but this should be f(0) = 0, f'(0) = 1.chapsticks said:Homework Statement
Solve the following differential equation:
f"(x)=sinx
x(0)=0, x'(0)=1
What's your question?chapsticks said:Homework Equations
x(0)=0, x'(0)=1
The Attempt at a Solution
f"(x)=sin(x)
integrate,
f'(x)=-cos(x)+C1
f'(0)=-cos(0)+C1=1 => C1=2
C1=2
f'(x)=-cos(x)+2
f(x)=-sin(x)+2x+C2
f(0)=-sin(0)+C2=0 => C2=0
=>
f(x)=-sin(x)+2x