What is the solution for 4y^'=(4xy+8x^2+2y^2)/(x^2) ?

The solution for 4y^'=(4xy+8x^2+2y^2)/(x^2) is y=2tan(ln(x)+c_{1})x

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4y^'=(4xy+8x^2+2y^2)/(x^2)

4 y^{^{'}} = \frac{4 x y + 8 x ^{2} + 2 y ^{2}}{x ^{2}}

y = 2 tan (ln (x) + c_{1}) x

Solution steps

4 y^{'} (x) = \frac{4 x y + 8 x ^{2} + 2 y ^{2}}{x ^{2}}

Solve using substitution:

y = 2 tan (ln (x) + c_{1}) x

y = 2 tan (ln (x) + c_{1}) x

y^{^{'}} + csc (y) = 0

\int \frac{e ^{3 x} + 1}{e ^{x} + 1} d x

x \to 3 lim (2 x^{3} - 4 x - 10)

9 x^{2} x + 7 x^{2}

(- 9, - 2), (3, - 1)

What is the solution for 4y^'=(4xy+8x^2+2y^2)/(x^2) ?
The solution for 4y^'=(4xy+8x^2+2y^2)/(x^2) is y=2tan(ln(x)+c_{1})x