(1+t^2)*y'+3ty=6t求常微分方程的通解

Question

(1+t^2)*y'+3ty=6t
求常微分方程的通解

Answer 1

求常微分方程的通解(1+t²)y'+3ty=6t
先求齐次方程(1+t²)y'+3ty=0的通解.
移项得(1+t²)(dy/dt)=-3ty；
分离变量得dy/y=[-3t/(1+t²)]dt；
积分之,得lny=-3∫[t/(1+t²)]dt=-(3/2)∫d(1+t²)/(1+t²)
=-(3/2)ln(1+t²)+lnC₁=ln[C₁/√(1+t²)³]
故得y=C₁/√(1+t²)³
下面用参数变易法求
将C₁换成t的函数u,得y=u/√(1+t²)³.(1)
将(1)的两边对t求导,得：
dy/dt=(du/dt)/√(1+t²)³-3ut/√(1+t²)⁵.(2)
将(1)和(2)代入原式得：
（1+t²)[(du/dt)/√(1+t²)³-3ut/√(1+t²)⁵]+3ut/√(1+t²)³=6t
化简得（1+t²)[(du/dt)/√(1+t²)³]=6t
即有(du/dt)/√(1+t²)=6t
分离变量得du=[6t√(1+t²)]dt
积分之,得u=∫[6t√(1+t²)]dt=3∫[d(1+t²)/√(1+t²)]=6√(1+t²)+C
代入(1)式,即得通解为y=[6√(1+t²)+C]/√(1+t²)³=6/√(1+t²)+C/√(1+t²)³.