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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int - 2 x e^{2 x} d x

\int - 2 x e^{2 x} d x = - e^{2 x} x + \frac{1}{2} e^{2 x} + c_{1}

y = - e^{2 x} x + \frac{1}{2} e^{2 x} + c_{1}

Apply initial conditions:

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

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

\frac{1}{s ^{1} ( s + 1 ) ^{3}}

f (x) = x^{2} - 4, - 3, 1

\int x x d x

\frac{d}{d t} (a cos (2 t) + b sin (2 t))

f (x) = (25 - 0.075 x) x, at x = 100

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