It is easy to check that for any value of c, the function y(x)=ce−2x+e−x y(x)=ce−2x+e−x is solution of equation y′+2y=e−x. y′+2y=e−x. find the value of cc for which the solution satisfies the initial condition y(−5)=3y(−5)=3.
Try this solution:1. if y(x)=Ce⁻²ˣ+e⁻ˣ and y(-5)=3, then it is possible to substitute the values of 'x' and 'y':C*e¹⁰+e⁵=3.2. it is possible to find the value of 'C':[tex]C= \frac{3-e^5}{e^{10}} [/tex]
