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

19.3 First-order equations: simple equations and separation of variables 543
Example19.10 Solve
dx
dt = t2 +1
x 2 +1
Solution Separating the variables and integrating we find
∫ ∫
x 2 +1dx= t 2 +1dt
Therefore,
x 3
3 +x = t3 3 +t+C
which is the general solution. Here we note that x has not been obtained explicitly in
terms oft, although we have found a relationship between x andt which satisfies the
differential equation. To obtain the value of x at any givent it would be necessary to
solve the cubic equation.
Sometimes,equationswhicharenotimmediatelyseparablecanbereducedtoseparable
formby anappropriatesubstitution asthe following example shows.
Example19.11 Bymeansofthe substitutionz = y , solve the equation
x
dy
dx = y2
x + y +1 (19.2)
2 x
Solution Ifz = y theny = zx. Because the solution,y, is a function ofxthe variablezdepends
x
uponxalso. The product rulegives dy
dx =z+xdz , so thatEquation (19.2) becomes
dx
z +x dz
dx =z2 +z+1
thatis,
x dz
dx =z2 +1
Thisnewequationhasindependentvariablexanddependentvariablez,andisseparable.
We find
∫
∫
dz dx
z 2 +1 = x
so thattan −1 z = ln|x| +C. WritingC = ln |D|wehave
tan −1 z = ln|x| +ln|D| = ln|Dx|
so that z = tan(ln|Dx|). Returning to the original variables we see that the general
solution is
y =zx =xtan(ln|Dx|)