What is the tangent of y-x^2y=2-2x^3 ?

The tangent of y-x^2y=2-2x^3 is y=(2(a_{0}^2+2a_{0}))/((a_{0)+1)^2}x+(4a_{0}+2)/((a_{0)+1)^2}

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tangent of y-x^2y=2-2x^3

tangent of

y - x^{2} y = 2 - 2 x^{3}

y = \frac{2 ( a _{0}^{2} + 2 a _{0} )}{( a _{0} + 1 ) ^{2}} x + \frac{4 a _{0} + 2}{( a _{0} + 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{2 ( a _{0}^{2} + a _{0} + 1 )}{a _{0} + 1})

Find the slope of

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

Compute the slope of

y - x^{2} y = 2 - 2 x^{3}

x = a_{0} : m = \frac{2 ( a _{0}^{2} + 2 a _{0} )}{( a _{0} + 1 ) ^{2}}

Find the line with slope m=

\frac{2 ( a _{0}^{2} + 2 a _{0} )}{( a _{0} + 1 ) ^{2}}

and passing through

(a_{0}, \frac{2 ( a _{0}^{2} + a _{0} + 1 )}{a _{0} + 1}) : y = \frac{2 ( a _{0}^{2} + 2 a _{0} )}{( a _{0} + 1 ) ^{2}} x + \frac{4 a _{0} + 2}{( a _{0} + 1 ) ^{2}}

y = \frac{2 ( a _{0}^{2} + 2 a _{0} )}{( a _{0} + 1 ) ^{2}} x + \frac{4 a _{0} + 2}{( a _{0} + 1 ) ^{2}}

\int \frac{7}{x ^{2} x ^{2} + 25} d x

\frac{d y}{d x} = (y - 1)^{2}

\frac{d y}{d t} = 6 y^{2}

f (x) = ln (x)

f (x) = 4 x + 1

What is the tangent of y-x^2y=2-2x^3 ?
The tangent of y-x^2y=2-2x^3 is y=(2(a_{0}^2+2a_{0}))/((a_{0)+1)^2}x+(4a_{0}+2)/((a_{0)+1)^2}