Find an explicit solution of the given initial-value problem. (1 + x4) dy + x (1 + 4y2) dx = 0, y (1) = 0

a solution is 1/2 * tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4

Step-by-step explanation:

for the equation

(1 + x⁴) dy + x * (1 + 4y²) dx = 0

(1 + x⁴) dy = - x * (1 + 4y²) dx

[1 / (1 + 4y²) ] dy = [-x / (1 + x⁴) ] dx

∫[1 / (1 + 4y²) ] dy = ∫[-x / (1 + x⁴) ] dx

now to solve each integral

I₁ = ∫[1 / (1 + 4y²) ] dy = 1/2 * tan⁻¹ (2*y) + C₁

I₂ = ∫[-x / (1 + x⁴) ] dx

for u = x² → du=x*dx

I₂ = ∫[-x / (1 + x⁴) ] dx = - ∫[1 / (1 + u²) ] du = - tan⁻¹ (u) + C₂ = - tan⁻¹ (x²) + C₂

then

1/2 * tan⁻¹ (2*y) = - tan⁻¹ (x²) + C

for y (x=1) = 0

1/2 * tan⁻¹ (2*0) = - tan⁻¹ (1²) + C

since tan⁻¹ (1²) for π/4 + π*N and tan⁻¹ (0) for π*N, we will choose for simplicity N=0. hen an explicit solution would be

1/2 * 0 = - π/4 + C

C = π/4

therefore

1/2 * tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4