What is the solution for y^'=(2x+1)/(x^2-3x+2),y(3)=0 ?

The solution for y^'=(2x+1)/(x^2-3x+2),y(3)=0 is y=-3ln(x-1)+ln(8(x-2)^5)

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y^'=(2x+1)/(x^2-3x+2),y(3)=0

y^{^{'}} = \frac{2 x + 1}{x ^{2} - 3 x + 2}, y (3) = 0

y = - 3 ln (x - 1) + ln (8 (x - 2)^{5})

Solution steps

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

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

then

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

y = \int \frac{2 x + 1}{x ^{2} - 3 x + 2} d x

\int \frac{2 x + 1}{x ^{2} - 3 x + 2} d x = - 3 ln (x - 1) + 5 ln (x - 2) + c_{1}

y = - 3 ln (x - 1) + 5 ln (x - 2) + c_{1}

Apply initial conditions:

y = - 3 ln (x - 1) + ln (8 (x - 2)^{5})

y = - 3 ln (x - 1) + ln (8 (x - 2)^{5})

x \frac{d y}{d x} = 1 + 3 y + y^{2}

\frac{d}{d x} (x cos (x) + 2 tan (x))

- x^{2} + 7 x + 2, at x = 3

y = 2 x^{3} - x^{2} + 1, (1, 2)

x \to 0 + lim (\frac{x}{x})

What is the solution for y^'=(2x+1)/(x^2-3x+2),y(3)=0 ?
The solution for y^'=(2x+1)/(x^2-3x+2),y(3)=0 is y=-3ln(x-1)+ln(8(x-2)^5)