Test for exactness. If exact, solve. If not, use an integrating factor as given or obtained by inspection or by the theorems in the text. Also, if an initial condition is given, find the corresponding particular solution.
Passo 1
We have the differential equation 2 x tan y
dx + sec² y
dy = 0. This is a first‐order ODE of the form M(x,y)
dx + N(x,y)
dy = 0, so we’ll test for exactness and, if needed, find an integrating factor.
dx + sec² y
dy = 0. This is a first‐order ODE of the form M(x,y)
dx + N(x,y)
dy = 0, so we’ll test for exactness and, if needed, find an integrating factor.
Passo 2
Compute ∂M/∂y and ∂N/∂x: M(x,y)=2 x tan y ⇒ ∂M/∂y = 2 x sec² y, N(x,y)=sec² y ⇒ ∂N/∂x = 0. Since ∂M/∂y ≠ ∂N/∂x, the equation is not exact.
Passo 3
Look for an integrating factor μ(x) depending only on x. We check (∂M/∂y − ∂N/∂x)/N = (2 x sec² y − 0)/sec² y = 2 x, which depends only on x. So μ(x) satisfies dμ/μ = 2 x
dx, giving μ(x) = e^{x²}.
dx, giving μ(x) = e^{x²}.
Passo 4
Multiply the original equation by μ(x)=e^{x²}: we get M̄ = 2 x e^{x²} tan y and N̄ = e^{x²} sec² y. Now ∂M̄/∂y = 2 x e^{x²} sec² y and ∂N̄/∂x = 2 x e^{x²} sec² y, so the new equation is exact.
Passo 5
Find a potential function φ(x,y) such that ∂φ/∂x = M̄. Integrate with respect to x: φ(x,y) = ∫2 x e^{x²} tan y
dx = e^{x²} tan y + h(y). Then ∂φ/∂y = e^{x²} sec² y + h’(y), which must equal N̄ = e^{x²} sec² y, so h’(y)=0 ⇒ h(y)=const. Thus φ(x,y)=e^{x²} tan y + C.
dx = e^{x²} tan y + h(y). Then ∂φ/∂y = e^{x²} sec² y + h’(y), which must equal N̄ = e^{x²} sec² y, so h’(y)=0 ⇒ h(y)=const. Thus φ(x,y)=e^{x²} tan y + C.
Passo 6
The general implicit solution is φ(x,y)=constant, i.e. e^{x²} tan y = C.
Resposta
e{x{2}} \tan y = C