Question

`y'cos(x)+ysin(x)=1` ??

Answer 1

`y = sin(x)+Ccos(x)`

Explanation:

Given: `y'cos(x)+ysin(x)=1`

Divide both sides of the equation by `cos(x)`:

`y' +tan(x)y = sec(x)`

A differential equation of the form, `y' + P(x)y= Q(x)`, is known to have the integrating factor of the form:

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

Multiplication by the integrating factor will cause the left side of the equation to be the derivative of `I(x)y`; this allows one to easily solve the equation.

By observation `P(x) = tan(x)`

`I(x) = e^(inttan(x)dx)`

`I(x) = e^(-ln|cos(x)|)`

`I(x) = sec(x)`

Multiply both sides of the equation by `I(x)`:

`y'sec(x) +tan(x)sec(x)y = sec^2(x)`

We know that the left side integrates to become `sec(x)y`:

`sec(x)y= intsec^2(x)dx`

`sec(x)y = tan(x)+C`

`y = tan(x)/sec(x) + C/sec(x)`

`y = sin(x)+Ccos(x)`