First Order ODE/y' = sin^2 (x - y + 1)

Theorem

The first order ODE:

$\dfrac {\d y} {\d x} = \map {\sin^2} {x - y + 1}^2$

has the general solution:

$\map \tan {x - y + 1} = x + C$

Proof

Make the substitution:

$z = x - y + 1$

Then from First Order ODE in form $y' = f (a x + b y + c)$ with $a = 1, b = - 1$:

\(\ds x\)

\(=\)

\(\ds \int \frac {\d z} {- \sin^2 z + 1}\)

\(\ds \)

\(=\)

\(\ds \int \frac {\d z} {\cos^2 z}\)

Sum of Squares of Sine and Cosine

\(\ds \)

\(=\)

\(\ds \int \sec^2 z \rd z\)

Secant is Reciprocal of Cosine

\(\ds \)

\(=\)

\(\ds \tan z + C_1\)

Primitive of $\sec^2 z$

\(\ds \leadsto \ \ \)

\(\ds \map \tan {x - y + 1}\)

\(=\)

\(\ds x + C\)

letting $C = - C_1$

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.7$: Homogeneous Equations: Problem $3 \ \text{(b)}$