If y = e^x.tan^-1x,prove that (1+x²)y₂ 2(1-x+x²)y₁ + (1-x²)y = 0

• y = e^x. tan^-1x
• y₁ = e^x (tan^-1x + 1/(1+x²) )
• y₁.(1+x²) = e^x + e^x.tan^-1x.(1+x²)
• y₁.(1+x²) = e^x + y.(1+x²)
• y₂.(1+x²) + y₁(2x) = e^x + y₁(1+x²) + y(2x)
• y₂.(1+x²) + y₁(2x)=(y₁.(1+x²) - y.(1+x²)) + y₁(1+x²) + y(2x)
• (1+x²)y₂ - 2(1-x+x²)y₁ + (1-x)²y= 0

This expression seems quite different from the one used asked to 'prove', but should be correct.