What is the solution for 2(ln(y))y^'-x^3y=0 ?

The solution for 2(ln(y))y^'-x^3y=0 is y=e^{(sqrt(x^4+c_{1)})/2},y=e^{-(sqrt(x^4+c_{1)})/2}

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2(ln(y))y^'-x^3y=0

2 (ln (y)) y^{^{'}} - x^{3} y = 0

y = e^{\frac{x ^{4} + c _{1}}{2}}, y = e^{- \frac{x ^{4} + c _{1}}{2}}

Solution steps

2 (ln (y)) y^{'} (x) - x^{3} y = 0

Solve separable ODE:

y = e^{\frac{x ^{4} + c _{1}}{2}}, y = e^{- \frac{x ^{4} + c _{1}}{2}}

y = e^{\frac{x ^{4} + c _{1}}{2}}, y = e^{- \frac{x ^{4} + c _{1}}{2}}

\int (- 3) d x

\int e^{s i n (θ)} d θ

f (x) = 4 + 3 x - 2 x^{2}, at x = 2

y = x (x^{3} + 3)^{5}

f (4 x^{2} - 2)

What is the solution for 2(ln(y))y^'-x^3y=0 ?
The solution for 2(ln(y))y^'-x^3y=0 is y=e^{(sqrt(x^4+c_{1)})/2},y=e^{-(sqrt(x^4+c_{1)})/2}