## - 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. $$x$$
$$\displaystyle \int \left(u {dy\over dx}+u P(x)y \right) dx=\int uQ(x) dx$$
$$\displaystyle \int x^{-3} {dy\over dx}-3x^{-4} y \:dx=\int 7x^{-3} dx$$
$$\displaystyle {y\over x^3} =-{7\over 2x^2}+c$$
$$\displaystyle y=-{7x\over 2}+cx^3$$

### 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
$$y=\sin⁡ x+\csc ⁡x$$

### 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.
$$y=e^{-4x}(e^x+2)$$