What is the solution for 2ty^'-4y=4t^4e^{t^2},y(1)=0 ?

The solution for 2ty^'-4y=4t^4e^{t^2},y(1)=0 is y=e^{t^2}t^2-et^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2ty^'-4y=4t^4e^{t^2},y(1)=0

2 t y^{^{'}} - 4 y = 4 t^{4} e^{t^{2}}, y (1) = 0

y = e^{t^{2}} t^{2} - e t^{2}

Solution steps

2 t y^{'} (t) - 4 y = 4 t^{4} e^{t^{2}}

Solve linear ODE:

y = e^{t^{2}} t^{2} - e t^{2}

y = e^{t^{2}} t^{2} - e t^{2}

\frac{d}{d x} (- \frac{6}{( 1 + x ) ^{4}})

\int \frac{1}{1 + 5 x ^{2}} d x

\int 4 sin^{3} (x co) s^{2} x d x

y = x + 225, (0, 15)

\int 2 x^{2} - 2 d x

What is the solution for 2ty^'-4y=4t^4e^{t^2},y(1)=0 ?
The solution for 2ty^'-4y=4t^4e^{t^2},y(1)=0 is y=e^{t^2}t^2-et^2