What is the solution for y^'=(6x-12)/(x^2),y(1)=20 ?

The solution for y^'=(6x-12)/(x^2),y(1)=20 is y=6ln(x)+(12)/x+8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=(6x-12)/(x^2),y(1)=20

y^{^{'}} = \frac{6 x - 12}{x ^{2}}, y (1) = 20

y = 6 ln (x) + \frac{12}{x} + 8

Solution steps

y^{'} (x) = \frac{6 x - 12}{x ^{2}}

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

then

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

y = \int \frac{6 x - 12}{x ^{2}} d x

\int \frac{6 x - 12}{x ^{2}} d x = 6 ln (x) + \frac{12}{x} + c_{1}

y = 6 ln (x) + \frac{12}{x} + c_{1}

Apply initial conditions:

y = 6 ln (x) + \frac{12}{x} + 8

y = 6 ln (x) + \frac{12}{x} + 8

\frac{d}{d b} (sec (b) \cdot csc (b) \cdot sin (b))

\int (\frac{( ln ( x ) ) ^{4}}{x}) d x

\int 2 x ln (x^{2}) d x

\int \frac{cos ^{3} ( x )}{sin ^{2} ( x )} d x

\int (\frac{7}{x} + \frac{15}{x ^{6}}) d x

What is the solution for y^'=(6x-12)/(x^2),y(1)=20 ?
The solution for y^'=(6x-12)/(x^2),y(1)=20 is y=6ln(x)+(12)/x+8