What is the solution for t^3(dy)/(dt)+3t^2y=2cos(t),y(pi)=0 ?

The solution for t^3(dy)/(dt)+3t^2y=2cos(t),y(pi)=0 is y=(2sin(t))/(t^3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

t^3(dy)/(dt)+3t^2y=2cos(t),y(pi)=0

t^{3} \frac{d y}{d t} + 3 t^{2} y = 2 cos (t), y (π) = 0

y = \frac{2 sin ( t )}{t ^{3}}

Solution steps

t^{3} \frac{d y}{d t} + 3 t^{2} y = 2 cos (t)

Solve linear ODE:

y = \frac{2 sin ( t )}{t ^{3}}

y = \frac{2 sin ( t )}{t ^{3}}

2 y^{^{'}} + y = 3 x^{2}

\int \frac{( 2 x - 3 ) ^{3}}{x} d x

\frac{d y}{d x} = y^{2} x

(x^{2} + 3)^{2} y = \frac{( 2 x ( y ^{2} + 1 ) ^{2} )}{( x ^{2} + 3 ) y ^{^{'}}} - (x^{4} + 6 x^{2} + 9) y

x y^{2} d x + (x^{2} y^{2} + x^{2} y) d y = 0

What is the solution for t^3(dy)/(dt)+3t^2y=2cos(t),y(pi)=0 ?
The solution for t^3(dy)/(dt)+3t^2y=2cos(t),y(pi)=0 is y=(2sin(t))/(t^3)