Question

Did I do this First order linear differential equation right?

y'-ytanx = 2xsinxcosx
Ok so I was told I.F. = `e^(intF(x))`
and I got y=cosx(sinx-xcosx)+C
Is this right? If not, please help me?

Answer 1

`y(x) = \frac{\tan (x)}{2}+\frac{1}{18} \sin (3 x)\sec(x)-\frac{2}{3} x \cos ^2(x)+C\sec x`

Explanation:

The integrating factor for a linear first order equation

`dy/dx +P(x) y= Q(x)`

is given by `e^{int P(x)dx}`

For our case `P(x) = -tan x` and `Q(x) = 2x sin x cos x`

Thus the integrating factor is

`e^{int (-tan x) dx} = e^{-ln sec x} = cos x`

Thus, we multiply both sides of the equation by `cos x`

This yields

`cos x dy/dx-y sin x = 2xsin xcos^2x`

The left hand side is

`cos x dy /dx +y d/dx (sin x) = d/dx(y cos x)`

and this means that

`y cos x = int 2 x sin s cos^2 x dx`

The integral can be evaluated easily by integration by parts to yield

`y cos x = \frac{\sin (x)}{2}+\frac{1}{18} \sin (3 x)-\frac{2}{3} x \cos ^3(x)+C`

and thus the solution is

`y(x) = \frac{\tan (x)}{2}+\frac{1}{18} \sin (3 x)\sec(x)-\frac{2}{3} x \cos ^2(x)+C\sec x`

Answer 2

`y = -2/3xcos^2x + 2/3tanx - 2/9sin^3xsecx + Csecx`

Explanation:

We have:

`y'-ytanx = 2xsinxcosx`

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

`dy/dx + P(x)y=Q(x)`

So, we can put the equation in standard form:

`y' - (tanx)y = 2xsinxcosx`

Then the integrating factor is given by;

`I = e^(int P(x) dx)`
`\ \ = exp(int \ -tanx) \ dx)`
`\ \ = exp(-lnsecx)`
`\ \ = exp(lncosx)`
`\ \ = cosx`

And if we multiply the DE [A] by this Integrating Factor, `I`, we will have a perfect product differential (in fact the original equation);

`:. cosxy' - cosx(tanx)y = cosx 2xsinxcosx` #

`:. cosxy' - sinxy = 2xsinxcos^2x`

`:. d/dx (cosx \ y) = 2xsinxcos^2x`

This is now separable, so by "separating the variables" we get:

`cosx \ y = int \ 2x \ sinx \ cos^2x \ dx`

Now, consider the integral:

`I = int \ 2x \ sinx \ cos^2x \ dx`

We can then apply Integration By Parts:

Let # { (u,=2x, => (du)/dx,=2), ((dv)/dx,=sinx \ cos^2x, => v,=-1/3cos^3x ) :}#

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

We have:

`int \ (2x)(sinx \ cos^2x) \ dx = (2x)(-1/3cos^3x) - int \ (-1/3cos^3x)(2) \ dx`

`:. I = -2/3xcos^3x + 2/3 \ int \ cos^3x \ dx`

So, next we must consider the integral:

`I_1 = int \ cos^3x \ dx`
`\ \ \ = int \ cosx \ cos^2x \ dx`
`\ \ \ = int \ cosx \ (1-sin^2x) \ dx`
`\ \ \ = int \ cosx \ dx - int \ cosx \ sin^2x \ dx`
`\ \ \ = sinx - 1/3sin^3x`

Combining these results, we have

`I = -2/3xcos^3x + 2/3 {sinx - 1/3sin^3x}`
`\ \ = -2/3xcos^3x + 2/3sinx - 2/9sin^3x`

So that, returning to the DE, we have:

`cosx \ y = -2/3xcos^3x + 2/3sinx - 2/9sin^3x + C`

`:. y = 1/cosx{-2/3xcos^3x + 2/3sinx - 2/9sin^3x + C}`

`:. y = -2/3xcos^2x + 2/3tanx - 2/9sin^3xsecx + Csecx`