What is the solution for t^2y^{''}+7ty^'+10y=0 ?

The solution for t^2y^{''}+7ty^'+10y=0 is y=(c_{1}cos(ln(t))+c_{2}sin(ln(t)))/(t^3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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})

x^{2} + 6

arcsin (\frac{x}{7})

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

What is the solution for t^2y^{''}+7ty^'+10y=0 ?
The solution for t^2y^{''}+7ty^'+10y=0 is y=(c_{1}cos(ln(t))+c_{2}sin(ln(t)))/(t^3)