y = ce^-2x + e^-x is solution of equation y' + 2y = e^-x
Find constant c so the solution satifies initial condition y(-2) = 5
2007-11-20 15:00:47 · 4 answers · asked by Gills 1 in Science & Mathematics ➔ Mathematics
dy/dx + 2y = e^(-x)
y (x) = C e^(-2x) + e^(-x)
y (- 2) = C e^4 + e^4 = 5
Ce^4 = 5 - e^4
C = (5 - e^4) / e^4
C = - 0.91
2007-11-21 03:11:05 · answer #1 · answered by Como 7 · 1⤊ 1⤋
y(x)=ce^(-2x)+e^(-x) using the initial condition y(-2)=5
y(-2)=ce^(4)+e^(2)=5
solving for c in the above equation we get
c=(5-e^2)/e^4
2007-11-24 12:10:42 · answer #2 · answered by curious 2 · 0⤊ 0⤋
c = (5-e^2)/e^4
2007-11-20 15:04:53 · answer #3 · answered by Mavis 5 · 0⤊ 0⤋
If y = ce^-2x + e^-x, then at (-2,5)
5 = c e^[(-2)(-2)] + e^[-(-2)],
which is now just algebra
2007-11-20 15:07:54 · answer #4 · answered by anobium625 6 · 0⤊ 1⤋