Question

Find the particular solution of the differential equation that satisfies the initial conditions. f ''(x) = 1/2 (e^x + e^−x), f(0) = 1, f '(0) = 0

Dont know where to begin

Answer 1

Answer: f(x) = cosh(x)

Explanation:

`f'(x)=\int\ f''{(x)} \, dx \\\\.\qquad =\int\ (1)/(2)(e^x+e^(-x))\ dx\\\\.\qquad =(1)/(2)\int (e^x+e^(-x))\ dx\\\\.\qquad =(1)/(2)\int 2\ cosh(x)\ dx\\\\.\qquad =(1)/(2)\cdot 2 \int cosh(x)\ dx\\\\.\qquad =\int cosh(x)\ dx\\\\.\qquad =sinh(x)+ C\\\\\text{Find the constant C by using the given condition that f'(0) = 0}\\0=sinh(0)+C\\0 = 0 + C\\0=C\\\\\text{So, f'(x) = sinh(x)}`

`f(x)=\int f'(x)\ dx\\\\.\qquad =\int sinh(x)\ dx\\\\.\qquad =cosh(x)+C\\\\\text{Find the constant C by using the given condition that f(0) = 1}\\1=cosh(0)+C\\1 = 1 + C\\0=C\\\\\text{So, f(x) = cosh(x)}`