What is the solution for y^{''}-4y^'+4y=((e^{2x}))/(x^2) ?

The solution for y^{''}-4y^'+4y=((e^{2x}))/(x^2) is y=c_{1}e^{2x}+c_{2}xe^{2x}-e^{2x}ln(x)-e^{2x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-4y^'+4y=((e^{2x}))/(x^2)

y^{^{''}} - 4 y^{^{'}} + 4 y = \frac{( e ^{2 x} )}{x ^{2}}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} - e^{2 x} ln (x) - e^{2 x}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = \frac{( e ^{2 x} )}{x ^{2}}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} - e^{2 x} ln (x) - e^{2 x}

y = c_{1} e^{2 x} + c_{2} x e^{2 x} - e^{2 x} ln (x) - e^{2 x}

e^{x}, e^{- 4}, 4, 0

\frac{440 t}{( t ^{2} + 7 ) ^{3}}

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

f (x) = e^{x}, at x = 2

y = 4^{- x}

What is the solution for y^{''}-4y^'+4y=((e^{2x}))/(x^2) ?
The solution for y^{''}-4y^'+4y=((e^{2x}))/(x^2) is y=c_{1}e^{2x}+c_{2}xe^{2x}-e^{2x}ln(x)-e^{2x}