Download Calculus II, Section 9.5, #10 Linear Equations Solve the differential

Calculus II, Section 9.5, #10
Linear Equations
Solve the differential equation.1
√
2xy ′ + y = 2 x
We want to solve the first-order linear differential equation (FOLDE)
√
2xy ′ + y = 2 x
The derivative in the equation is y ′ , so the independent variable is x. This would be more clear if the author
dy
had written y ′ (x) or dx
, but since there are no other parameters in the equation, the meaning is (mostly)
clear.
√
Before we start, note that
x imposes the condition x ≥ 0 upon the independent variable x.
We want to write the FOLDE in the form
dy
+ P (x) y = Q(x)
dx
√
2xy ′ + y = 2 x
√
2 x
1
y=
y′ +
2x
2x
1
1
′
y= √
y +
2x
x
Before we continue, note that
√1
x
now imposes the condition x > 0 upon the independent variable x.
The integrating factor is
int. factor = e
R
1
= e2
1
2x
R
dx
1
x
dx
1
= e 2 ln |x|
and since x > 0,
1
= e 2 ln x
1
= eln x 2
1
= x2
√
= x
Multiplying our FOLDE by this integrating factor, we get
√ ′
xy +
1 Stewart,
√
√
x
x
y= √
2x
x
Calculus, Early Transcendentals, p. 625, #10.
Calculus II
Linear Equations
√ ′
1
xy + √ y = 1
2 x
So
√ ′
xy = 1
and we integrate
Z
Z
√ ′
x y dx = 1dx
√
xy = x+C
x+C
y= √
x
√
Thus the general solution to the FOLDE 2xy ′ + y = 2 x is y =
x+C
√ .
x