t^2y^{''}+7ty^'+10y=0

解答

t^{2} y^{^{''}} + 7 t y^{^{'}} + 10 y = 0

y = \frac{c _{1} cos ( ln ( t ) ) + c _{2} sin ( ln ( t ) )}{t ^{3}}

求解步骤

t^{2} y^{''} (t) + 7 t y^{'} (t) + 10 y = 0

求解线性方程:

y = \frac{c _{1} cos ( ln ( t ) ) + c _{2} sin ( ln ( t ) )}{t ^{3}}

y = \frac{c _{1} cos ( ln ( t ) ) + c _{2} sin ( ln ( t ) )}{t ^{3}}

\int tan (4 x) sec^{5} (4 x) d x

x \to 7 - lim (\frac{1}{x - 7})

d er i v a t i v e x^{2} + 6

t an g e n t arcsin (\frac{x}{7})

y^{^{'}} = \frac{x ^{2} + 2 \cdot y ^{2}}{x \cdot y}