What is the solution for x^'=(x)^{4(1)+2}(ln(t)+(0+2)te^t) ?

The solution for x^'=(x)^{4(1)+2}(ln(t)+(0+2)te^t) is x=-\sqrt[5]{1/(5(tln(t)-t+2e^tt-2e^t+c_{1))}}

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x^'=(x)^{4(1)+2}(ln(t)+(0+2)te^t)

x^{^{'}} = (x)^{4 (1) + 2} (ln (t) + (0 + 2) t e^{t})

Solution steps

x^{'} (t) = (x)^{4 (1) + 2} (ln (t) + (0 + 2) t e^{t})

Solve separable ODE:

\frac{x}{y} d x + \frac{y}{x} d y = 0

x^{^{'}} = e^{x} \frac{1}{t}

(3 x - 5) d x + (y + b) d y = 0

\frac{d y}{d x} = 3 y^{\frac{2}{3}}, y (2) = 0

y y^{^{'}} = e^{2 x} + 1

What is the solution for x^'=(x)^{4(1)+2}(ln(t)+(0+2)te^t) ?
The solution for x^'=(x)^{4(1)+2}(ln(t)+(0+2)te^t) is x=-\sqrt[5]{1/(5(tln(t)-t+2e^tt-2e^t+c_{1))}}