t^2y^'+y=sin(t),y(1)=0

解答

t^{2} y^{^{'}} + y = sin (t), y (1) = 0

y = e^{\frac{1}{t}} \cdot \int \frac{e ^{- \frac{1}{t}} sin ( t )}{t ^{2}} d t

求解步骤

t^{2} y^{'} (t) + y = sin (t)

求解线性方程:

y = e^{\frac{1}{t}} \cdot \int \frac{e ^{- \frac{1}{t}} sin ( t )}{t ^{2}} d t

y = e^{\frac{1}{t}} \cdot \int \frac{e ^{- \frac{1}{t}} sin ( t )}{t ^{2}} d t