082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

542 Chapter 19 Ordinary differential equations I
Example19.8 Solve dy
dx = e−x
y .
Solution Here f (x) = e −x andg(y) = 1 . Multiplication through byyyields
y
y dy
dx = e−x
Integration of both sides with respect toxgives
∫ ∫
ydy= e −x dx
so that
y 2
2 =−e−x +C
Note that the constants arising from the two integrals have been combined to give a
single constant,C. Finally we can rearrange this expression togiveyinterms ofx:
that is,
y 2 =−2e −x +2C
y=± √ D−2e −x
whereD = 2C.
It is important to stress that the constant of integration must be inserted at the stage at
which the integration is actually carried out, and not simply added to the answer at the
end.
Example19.9 Solve dy
dx = 3x2 e −y subjecttoy(0) = 1.
Solution Hereg(y) = e −y and f (x) = 3x 2 . Separating the variables and integrating wefind
∫ ∫
e y dy= 3x 2 dx
so thate y =x 3 +C. Imposing the initial conditiony(0) = 1 wefind
e 1 =(0) 3 +C
so thatC = e.Therefore,
e y =x 3 +e
Note that since the exponential function is always positive, the solution will be valid
only forx 3 +e > 0.Taking natural logarithms gives the particular solution explicitly:
y =ln(x 3 +e)