Question

What is the general solution of the differential equation `dy/dx +y = y^2(cosx-sinx)`?

Answer 1

`y = 1/( C e^x - sin x)`

Explanation:

`y' + y = y^2 (cos x - sin x)`

`(y')/y^2 + 1/y = cos x - sin x`

Let `v(x) = 1/(y(x))`, such that `v' = - (y')/y^2`

`implies - v' + v = cos x - sin x`

`implies v' - v = - cos x + sin x`

From here, we grab the Integrating Factor, `\lambda(x)`:

`lambda(x) = e^(- int dx) = e^(-x)`

`implies e^(-x) ( v' - v) = e^(-x) ( - cos x + sin x)`

`implies v e^(-x) = int dx qquad e^(-x) ( - cos x + sin x)`

`qquad = int dx qquad d/dx ( -e^(-x) sin x)`

`implies v e^(-x) = -e^(-x) sin x + C`

From there, we remember that: `v = 1/y`.

Therefore:

`y = 1/( C e^x - sin x)`

Answer 2

The solution to the DE

`dy/dx +y = y^2(cosx-sinx)`

is:

`y = 1/(Ce^x - sin x)`

Explanation:

We have:

`dy/dx +y = y^2(cosx-sinx) \ \ \ \ \ ..... [1]`

This is a First Order Bernoulli differential equation of the form:

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

The technique to solve such an equation is to use a substitution to reduce the equation to a linear differential equation. We can do this by letting `v = y^(-1)`

`v=y^(-1) <=> y=1/v`

Differentiating we get:

`\ \ \ \ \ (dv)/dy = -1/y^2`
`\ \ \ \ \ (dy)/dx = (dy)/(dv)*(dv)/dx \ \ \ ` (chain rule)
`:. (dy)/dx = -y^2 \ (dv)/dx`
`:. (dy)/dx = -1/v^2 \ (dv)/dx`

Substituting into the DE [1] gives us:

`-1/v^2 \ (dv)/dx +1/v = 1/v^2(cosx-sinx)`
`:. -(dv)/dx + v = cosx-sinx`
`:. (dv)/dx - v = sinx-cosx \ \ \ \ \ [2]`

And so we have reduced our non-linear DE to a First Order Linear Differential equation of the form

`(dv)/dx + P(x)v = Q(x)`

We can sole this equation by using an Integrating Factor to convert the equation to a the perfect differential of a product as follows:

Let `I = exp( \ int \ P(x) \ dx \ )`
`\ \ \ \ \ \ \ = exp( \ int \ -1 \ dx \ )`
`\ \ \ \ \ \ \ = exp( -x )`
`\ \ \ \ \ \ \ = e^( -x )`

Multiply the DE [2] by the IF:

`e^( -x )(dv)/dx - ve^( -x ) = e^( -x )(sinx-cosx)`
`d/dx \ (ve^( -x )) = e^( -x )(sinx-cosx)`

This is a simple separable DE so we can separate the variables to get:

`ve^( -x ) = int \ e^( -x )(sinx-cosx) dx`

To save space I will just quote the result for the integral on the RHS, but this can easily be derived using Integration by Parts, so we get:

`ve^( -x ) = -e^( -x )sinx + C`

Restoring the earlier substitution the gives:

`y^(-1) e^( -x ) = e^( -x )sinx + C`
`:. \ y^(-1) = -sinx + Ce^x`
`:. \ \ \ \ 1/y = Ce^x - sin x`
`:. \ \ \ \ \ y = 1/(Ce^x - sin x)`