y^{''}+y=x^2+1,y(0)=2,y(1)=0

解答

y^{^{''}} + y = x^{2} + 1, y (0) = 2, y (1) = 0

y = 3 cos (x) - \frac{3 cos ( 1 ) sin ( x )}{sin ( 1 )} + x^{2} - 1

求解步骤

y^{''} (x) + y = x^{2} + 1

求解线性方程:

y = 3 cos (x) - \frac{3 cos ( 1 ) sin ( x )}{sin ( 1 )} + x^{2} - 1

y = 3 cos (x) - \frac{3 cos ( 1 ) sin ( x )}{sin ( 1 )} + x^{2} - 1

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 4 y = 3 x - 4

\frac{d ^{2} x}{d t ^{2}} + 9 x = e^{t} + sin (t)

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{2} e^{2 t} + e^{2 t}

y^{^{''}} - y = (1 - e^{2} x)

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 9 y = 2 sin (4 x)