What is the solution for 3yy^'=7x ?

The solution for 3yy^'=7x is y=sqrt((7x^2+c_{1))/3},y=-sqrt((7x^2+c_{1))/3}

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3yy^'=7x

3 y y^{^{'}} = 7 x

y = \frac{7 x ^{2} + c _{1}}{3}, y = - \frac{7 x ^{2} + c _{1}}{3}

Solution steps

3 y y^{'} (x) = 7 x

Solve separable ODE:

y = \frac{7 x ^{2} + c _{1}}{3}, y = - \frac{7 x ^{2} + c _{1}}{3}

y = \frac{7 x ^{2} + c _{1}}{3}, y = - \frac{7 x ^{2} + c _{1}}{3}

8 + x - 7

y^{^{''}} - x = 0

x \to 1 lim (ln (x^{4} - 3 x + 10))

x \to - 3 + lim (\frac{x ^{2}}{x + 3})

\int arctan (3 x) d x

What is the solution for 3yy^'=7x ?
The solution for 3yy^'=7x is y=sqrt((7x^2+c_{1))/3},y=-sqrt((7x^2+c_{1))/3}