What is expand (e^{x^2}y= 1/2 e^{x^2}+c) 1/(e^{x^2)} ?

The solution to expand (e^{x^2}y= 1/2 e^{x^2}+c) 1/(e^{x^2)} is y= 1/2 e^{x^2}+c/(e^{x^2)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand (e^{x^2}y= 1/2 e^{x^2}+c) 1/(e^{x^2)}

expand

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

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

Solution steps

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

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

Apply the distributive law:

a (b + c) = ab + a c

a = \frac{1}{e ^{x^{2}}}, b = e^{x^{2}} y = \frac{1}{2} e^{x^{2}}, c = c

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

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

Simplify

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

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

\frac{3 ( 7 4 ^{\frac{2}{3}} - 5 7 ^{\frac{2}{3}} )}{4}

\frac{3 ( b + 4 )}{b + 3}

- \frac{10}{( 4 + 3 x ) ^{2}}

(\frac{y ^{3}}{2} - \frac{2}{y ^{3}})^{2}

x (2 ln (x) + 1)

What is expand (e^{x^2}y= 1/2 e^{x^2}+c) 1/(e^{x^2)} ?
The solution to expand (e^{x^2}y= 1/2 e^{x^2}+c) 1/(e^{x^2)} is y= 1/2 e^{x^2}+c/(e^{x^2)}