What is the solution for x^2y^{''}-3xy^'+4y=x,y(1)=0,y^'(1)=0 ?

The solution for x^2y^{''}-3xy^'+4y=x,y(1)=0,y^'(1)=0 is y=-x^2+x^2ln(x)+x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2y^{''}-3xy^'+4y=x,y(1)=0,y^'(1)=0

x^{2} y^{^{''}} - 3 x y^{^{'}} + 4 y = x, y (1) = 0, y^{^{'}} (1) = 0

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

Solution steps

x^{2} y^{''} (x) - 3 x y^{'} (x) + 4 y = x

Solve linear ODE:

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

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

\frac{d y}{d x} = \frac{x}{x ^{2} + 64}

n = 1 \sum \infty n! (2 x - 1)^{n}

\frac{d x}{d t} = - 8

\int \frac{x ^{4} + 5 x ^{3} - 3}{x} d x

\int (6 x y + 4 x) d y

What is the solution for x^2y^{''}-3xy^'+4y=x,y(1)=0,y^'(1)=0 ?
The solution for x^2y^{''}-3xy^'+4y=x,y(1)=0,y^'(1)=0 is y=-x^2+x^2ln(x)+x