y
=
y
(
x
)
,
{\displaystyle y=y(x),}
and
y
{\displaystyle y}
and its derivative are both of the first degree.
d
y
d
x
+
P
(
x
)
y
=
Q
(
x
)
{\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}+P(x)y=Q(x)}
To solve this equation, we use an integrating factor
e
∫
P
(
x
)
d
x
.
{\displaystyle e^{\int P(x)\mathrm {d} x}.}
We will provide an example and show that this integrating factor makes the above equation exact, as intended.
Example
Solve the following equation. Because the degree of y {\displaystyle y} y and its derivative are both 1, this equation is linear. d y d x + 2 x y = 4 sin x {\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}+{\frac {2}{x}}y=4\sin x} {\frac {{\mathrm {d}}y}{{\mathrm {d}}x}}+{\frac {2}{x}}y=4\sin x
Find the integrating factor. e ∫ P ( x ) d x = e ∫ 2 / x d x = e 2 ln | x | = x 2 {\displaystyle {\begin{aligned}e^{\int P(x)\mathrm {d} x}&=e^{\int 2/x\mathrm {d} x}\\&=e^{2\ln |x|}\\&=x^{2}\end{aligned}}} {\begin{aligned}e^{{\int P(x){\mathrm {d}}x}}&=e^{{\int 2/x{\mathrm {d}}x}}\\&=e^{{2\ln |x|}}\\&=x^{{2}}\end{aligned}}
Rewrite the equation in Pfaffian form and multiply by the integrating factor. We can confirm that this is an exact differential equation by doing the partial derivatives. ( 2 x y − 4 x 2 sin x ) d x + x 2 d y = 0 {\displaystyle (2xy-4x^{2}\sin x)\mathrm {d} x+x^{2}\mathrm {d} y=0} (2xy-4x^{{2}}\sin x){\mathrm {d}}x+x^{{2}}{\mathrm {d}}y=0
Solve this equation using any means possible. We write f {\displaystyle f} f as a solution to the differential equation. f = ∫ Q d y = x 2 y + R ( x ) {\displaystyle f=\int Q\mathrm {d} y=x^{2}y+R(x)} f=\int Q{\mathrm {d}}y=x^{{2}}y+R(x) ∂ f ∂ x = P = 2 x y + d R d x {\displaystyle {\frac {\partial f}{\partial x}}=P=2xy+{\frac {\mathrm {d} R}{\mathrm {d} x}}} {\frac {\partial f}{\partial x}}=P=2xy+{\frac {{\mathrm {d}}R}{{\mathrm {d}}x}} R ( x ) = ( 4 x 2 − 8 ) cos x − 8 x sin x {\displaystyle R(x)=(4x^{2}-8)\cos x-8x\sin x} R(x)=(4x^{{2}}-8)\cos x-8x\sin x x 2 y + ( 4 x 2 − 8 ) cos x − 8 x sin x = C {\displaystyle x^{2}y+(4x^{2}-8)\cos x-8x\sin x=C} x^{{2}}y+(4x^{{2}}-8)\cos x-8x\sin x=C
Derivation of the Integrating Factor
Rewrite the linear differential equation in Pfaffian form. ( P ( x ) y − Q ( x ) ) d x + d y = 0 {\displaystyle (P(x)y-Q(x))\mathrm {d} x+\mathrm {d} y=0} (P(x)y-Q(x)){\mathrm {d}}x+{\mathrm {d}}y=0
Consider an integrating factor μ ( x ) {\displaystyle \mu (x)} \mu (x). This integrating factor is such that multiplying the above equation by it makes the equation exact. ( μ P ( x ) y − μ Q ( x ) ) d x + μ ( x ) d y = 0 {\displaystyle (\mu P(x)y-\mu Q(x))\mathrm {d} x+\mu (x)\mathrm {d} y=0} (\mu P(x)y-\mu Q(x)){\mathrm {d}}x+\mu (x){\mathrm {d}}y=0
Invoke the necessary and sufficient condition for exactness. To be exact, the coefficients of the differentials must satisfy Clariaut's theorem. ∂ ∂ y ( μ P ( x ) y − μ Q ( x ) ) = ∂ μ ∂ x {\displaystyle {\frac {\partial }{\partial y}}(\mu P(x)y-\mu Q(x))={\frac {\partial \mu }{\partial x}}} {\frac {\partial }{\partial y}}(\mu P(x)y-\mu Q(x))={\frac {\partial \mu }{\partial x}}
Simplify the resulting expression. We recognize that P ( x ) , {\displaystyle P(x),} P(x), Q ( x ) , {\displaystyle Q(x),} Q(x), and μ {\displaystyle \mu } \mu are all functions of x {\displaystyle x} x only. μ P ( x ) = d μ d x {\displaystyle \mu P(x)={\frac {\mathrm {d} \mu }{\mathrm {d} x}}} \mu P(x)={\frac {{\mathrm {d}}\mu }{{\mathrm {d}}x}}
Separate variables and integrate to solve for μ {\displaystyle \mu } \mu . 1 μ d μ = P ( x ) d x ln | μ | = ∫ P ( x ) d x μ = e ∫ P ( x ) d x ( QED. ) {\displaystyle {\begin{aligned}{\frac {1}{\mu }}\mathrm {d} \mu &=P(x)\mathrm {d} x\\\ln |\mu |&=\int P(x)\mathrm {d} x\\\mu &=e^{\int P(x)\mathrm {d} x}\ \ \ \ \ \ ({\text{QED.}})\end{aligned}}} {\begin{aligned}{\frac {1}{\mu }}{\mathrm {d}}\mu &=P(x){\mathrm {d}}x\\\ln |\mu |&=\int P(x){\mathrm {d}}x\\\mu &=e^{{\int P(x){\mathrm {d}}x}}\ \ \ \ \ \ ({\text{QED.}})\end{aligned}}
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