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

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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

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

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

\frac{\partial}{\partial x} ((\frac{x + 2}{y + 1})^{\frac{1}{2}})

l a pl a ce t r an s f or m e^{a s}

n \to \infty lim (- 3 + 4 n^{5})

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