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

The solution for y^{''}-y^'=xe(x),y(0)=2,y^'(0)=1 is y=(2-(1-2e))e^x+1-2e-(ex^3)/3-ex^2-2ex

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y^'=xe(x),y(0)=2,y^'(0)=1

y^{^{''}} - y^{^{'}} = x e (x), y (0) = 2, y^{^{'}} (0) = 1

y = (2 - (1 - 2 e)) e^{x} + 1 - 2 e - \frac{e x ^{3}}{3} - e x^{2} - 2 e x

Solution steps

y^{''} (x) - y^{'} (x) = x e (x)

Solve linear ODE:

y = (2 - (1 - 2 e)) e^{x} + 1 - 2 e - \frac{e x ^{3}}{3} - e x^{2} - 2 e x

y = (2 - (1 - 2 e)) e^{x} + 1 - 2 e - \frac{e x ^{3}}{3} - e x^{2} - 2 e x

x^{^{''}} + x^{^{'}} + 12 x = 40 cos (3 t)

y^{^{''}} - 3 y^{^{'}} = e^{3 x}

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 e^{t}

y^{^{'''}} - y^{^{''}} + y^{^{'}} - y = 9 e^{- x}

y^{^{''}} + 4 y = - 20 sin (2 t)

What is the solution for y^{''}-y^'=xe(x),y(0)=2,y^'(0)=1 ?
The solution for y^{''}-y^'=xe(x),y(0)=2,y^'(0)=1 is y=(2-(1-2e))e^x+1-2e-(ex^3)/3-ex^2-2ex