Question

What is the particular solution of the differential equation `y' + y tanx = sin(2x)` where `y(0)=1`?

Answer 1

`y = -2cos^2(x)+3cos(x)`

Explanation:

The given equation,

`y'+ytan(x)=sin(2x)`

, is of the form:

`y' + P(x)y = Q(x)`

Where `P(x) = tan(x), and Q(x) = sin(2x)`

It is known that the integrating factor is:

`mu(x) = e^(intP(x)dx)`

`inttan(x)dx = log(sec(x))`

`mu(x) = e^log(sec(x)) = sec(x)`

Multiply the given equation by `mu(x)`:

`y'sec(x)+tan(x)sec(x)y=sin(2x)/sec(x)`

We know that the left side integrates to `mu(x)y` and we are left with the task of integrating the right side:

`sec(x)y = intsin(2x)sec(x)dx`

`sec(x)y = -2cos(x)+C`

Multiply both side by `cos(x)`

`y = -2cos^2(x)+Ccos(x)`

Use the boundary condition to find the value of C:

`1 = -2cos^2(0)+Ccos(0)`

`C = 3`

Answer 2

`y = -2cos^2x + 3cosx`

Explanation:

We have

`y' + y tanx = sin(2x)`

Which can be written:

`y' + tanx \ y = sin(2x)`

This is a first order linear ordinary differential equation of the form:

`(d zeta)/dx + P(x) zeta = Q(x)`

We solve this using an Integrating Factor

`I = exp( \ int \ P(x) \ dx )`
`\ \ = exp( int \ (tanx) \ dx )`
`\ \ = exp( ln |secx| )`
`\ \ = exp( ln secx )`
`\ \ = secx`

And if we multiply the DE by this Integrating Factor, `I`, we will have a perfect product differential;

`secx y' + y secxtanx = secx sin(2x)`
`:. d/dx(ysecx) = secx sin(2x)`

Which is now a separable DE, so we can "separate the variables" to get:

`ysec x = int \ secx sin(2x) \ dx`
`" " = int \ 1/cosx (2sinxcosx) \ dx`
`" " = 2 \ int \ sinx \ dx`

And, Integrating we get:

`ysecx = -2cosx + C`

`:. y/cosx = -2cosx + C`
`:. y = -2cos^2x + Ccosx`

Which is the general Solution. Then using `y(0)=1` we can find `C` as:

`1 = -2cos^2 0 + Ccos0 => 1 = -2 + C => C= 3`

Hence

`:. y = -2cos^2x + 3cosx`