- Integrating Factor

Linear differential equations of the form

\(\displaystyle {dy\over dx}+P(x)y=Q(x)\)
can be solved by multiplying throughout by the Integrating Factor, \(e^{∫P(x)}\).

LESSON 1

Solve the following differential equations

\(\displaystyle x {dy\over dx}-3y=7x\)
SOLUTION
  • Write the DE in the form
\(\displaystyle {dy\over dx}+P(x)y=Q(x)\)
\(\displaystyle ⟹{dy\over dx}-{3\over x} y=7\)
  • Calculate the I.F using
I.F\(=e^{∫P(x) dx}\)
\(\displaystyle P(x)=-{3\over x}\)
I.F\(\displaystyle =e^{∫-{3\over x}dx}=e^{-3 \ln ⁡x}=e^{\ln⁡ x^{-3}}=x^{-3}\)
  • Multiply each term in the equation by I.F
\(\displaystyle x^{-3} {dy\over dx}-3x^{-4} y=7x^{-3}\)
L.H.S resembles the product rule where I.F is actually \(u\)
  • Integrate both sides of the equation w.r.t. \(x\)
\(\displaystyle \int \left(u {dy\over dx}+u P(x)y \right) dx=\int uQ(x) dx\)
\(\displaystyle \int x^{-3} {dy\over dx}-3x^{-4} y \:dx=\int 7x^{-3} dx\)
\(\displaystyle {y\over x^3} =-{7\over 2x^2}+c\)
\(\displaystyle y=-{7x\over 2}+cx^3\)

VIDEO SOLUTION


LESSON 2

Solve the differential equation

\(\displaystyle {dy\over dx}+(\cot ⁡x)y=2 \cos⁡ x\)
given that \(y=2\) when \(\displaystyle x={π\over 2}\).
SOLUTION
\(\displaystyle {dy\over dx}+(\cot ⁡x)y=2 \cos ⁡x\)
  • Determine \(P(x)\)
\(P(x)=\cot ⁡x\)
  • Determine the Integrating Factor
\(\displaystyle \int \cot ⁡x\: dx=∫{\cos ⁡x\over \sin ⁡x} dx =\ln⁡\sin x\)
I.F\(\displaystyle =e^{\ln \sin x}=\sin ⁡x\)
  • Multiply throughout by the Integrating Factor
\(\displaystyle \sin ⁡x {dy\over dx}+\sin ⁡x.{\cos ⁡x\over \sin ⁡x}.y=2 \sin ⁡x \cos ⁡x\)
\(\displaystyle \int \sin ⁡x {dy\over dx}+(\cos ⁡x)y \:dx=2\int \cos ⁡x \sin ⁡x \:dx\)
  • Integrate both sides
\(y \sin⁡x=\sin^2⁡x+K\)
\(y=\sin ⁡x+K \csc⁡ x\)
  • Use given values to determine the value of the unknown constant.
\(y=2\) when \(\displaystyle x={π\over 2}\)
\(\displaystyle 2=\sin⁡ \left({π\over2}\right)+{K\over \sin ⁡\left({π\over2}\right)}\)
\(\displaystyle 2=1+{K\over1}\)
\(K=1\)
  • Substitute determined value for the unknown into the general solution
\(y=\sin⁡ x+\csc ⁡x\)

VIDEO LESSON


LESSON 3

Determine the particular solution of the differential equation
\(\displaystyle dy+4y\:dx=e^{-3x}\: dx\)

given that \(y=3\) when \(x=0\).

SOLUTION
  • Rewrite the equation in the form \(\displaystyle {dy\over dx}+P(x)y=Q(x)\)
\(dy+4ydx=e^{-3x}\: dx\)
\(\displaystyle {dy\over dx}+4y=e^{-3x}\)
  • Determine the Integrating Factor.
I.F\(\displaystyle =e^{\int 4\: dx}=e^{4x}\)
  • Multiply throughout by the Integrating Factor.
\(\displaystyle e^{4x} {dy\over dx}+4e^{4x} y=e^{-3x} e^{4x}\)
\(\displaystyle \int e^{4x} {dy\over dx}+4e^{4x} y\: dx=\int e^x \:dx\)
  • Integrate both sides w.r.t \(x\)
\(ye^{4x}=e^x+c\)
\(y=e^{-4x} (e^x+c)\)
  • Use the given values to determine the value of the
When \(y=3,\:x=0\)
\(3=e^{4(0)}(e^0+c)\)
\(2=c\)
  • Substitute value of the unknown constant into the general solution.
\(y=e^{-4x}(e^x+2)\)

VIDEO LESSON