Question

怎么用常数变易法求y'-y=2exarctanx的通解?

Answer 1

说明：原题应该是y'-y=2e^x*arctanx.

∵齐次方程y'-y=0的通解是y=Ce^x (C是积分常数)

∴根据常数变易法,设原微分方程的解为y=C(x)e^x (C(x)是关于x的函数)

∵y=C'(x)e^x+C(x)e^x

代入原方程得C'(x)e^x+C(x)e^x-C(x)e^x=2e^x*arctanx

==>C'(x)=2arctanx

==>C(x)=∫2arctanxdx=2x*arctanx-ln(1+x²)+C (C是积分常数)

∴y=C(x)e^x=(2x*arctanx-ln(1+x²)+C)e^x

故原微分方程的通解是y=(2x*arctanx-ln(1+x²)+C)e^x (C是积分常数).