In case an equation is in the form y\' = f(ax + by + c), i.e., the RHS is a line

Question

In case an equation is in the form y' = f(ax + by + c), i.e., the RHS is a linear function of x and y We will use the substitution v = ax + by + c to find an implicit general solution. The right hand side of the following first order problem is a linear function of x and y. Use the substitution v = x - y to solve the initial value problem. y' = sin^2(x - y), y(1) = 1 We obtain the following separable equation in the variables x and v: Solving this equation and transforming back to the variables x and y we arrive at the implicit solution Finally we obtain the explicit solution of the initial value problem as

Explanation / Answer

y' = sin2(x - y) , y(1) = 1

let x - y = v

differentiate ==> 1 - y' = v'

==> y' = 1 - v'

==> y' = sin2(x - y)

==> 1 - v' = sin2(v)

==> v' = 1 - sin2v

==> v' = cos2v

==> v'/ cos2v = 1

==> sec2v v' = 1

Integrating on both sides with respect to x

==> sec2v (dv/dx) dx = 1 dx

==> sec2v dv = 1 dx

==> tanv = x + c

==> v = tan-1(x + c)

substitute back v = x - y

==> x - y = tan-1(x + c)          ===> tan(x - y) = x + c

==> y = x - tan-1(x + c)

given y(1) = 1

==> 1 = 1 - tan-1(1 + c)

==> tan-1(1 + c) = 0

==> 1 + c = 0

==> c = -1

==> y = x - tan-1(x - 1)

Hence solution to given differential equation is y = x - tan-1(x - 1)

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