What is the solution for (2x+3)^4=3y^4(dy)/(dx) ?

The solution for (2x+3)^4=3y^4(dy)/(dx) is y=\sqrt[5]{(16x^5)/3+40x^4+120x^3+180x^2+135x-c_{1}}

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(2x+3)^4=3y^4(dy)/(dx)

(2 x + 3)^{4} = 3 y^{4} \frac{d y}{d x}

Solution steps

(2 x + 3)^{4} = 3 y^{4} \frac{d y}{d x}

Solve separable ODE:

\frac{d y}{d x} = - \frac{4}{6} x^{8} y^{6}

(2 x - 1) y^{^{'}} - 2 y = 0 \frac{1 - 4 x}{x ^{2}}

\frac{d x}{d t} = k (1 - x)

\frac{d y}{d x} + y = y x e^{x + 7}

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

What is the solution for (2x+3)^4=3y^4(dy)/(dx) ?
The solution for (2x+3)^4=3y^4(dy)/(dx) is y=\sqrt[5]{(16x^5)/3+40x^4+120x^3+180x^2+135x-c_{1}}