Chapter 1 - First-Order Differential Equations - 1.6 First-Order Linear Differential Equations - Problems - Page 59: 19

\[y(x)=\cos hx\]

$(y-e^x)dx+dy=0$ $dy=(e^x-y)dx$ $\frac{dy}{dx}+y=e^x$ ___(1) This is linear differential equation Integrating factor:- $I(x)=e^{\int 1dx}=e^x$ Multiply (1) by $e^x$ $e^x\frac{dy}{dx} +e^xy=e^{2x}$ $\frac{d}{dx}(ye^x)=e^{2x}$ Integrating, $ye^x=\int e^{2x}dx+C$ $ye^x=\frac{e^{2x}}{2}+C$ ___(2) Using initial condition $y(0)=1$ $1=\frac{1}{2}+C\;\Rightarrow C=\frac{1}{2}$ From (2) $2ye^x=e^{2x}+1$ $y=\frac{e^x+e^{-x}}{2}=\cos hx$