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

The solution for (dy)/(dx)(x)=y+(x^2-3)^2 is y=-x^4-6x^2+c_{1}x+(4x^4)/3-9

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

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

y = - x^{4} - 6 x^{2} + c_{1} x + \frac{4 x ^{4}}{3} - 9

Solution steps

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

Solve linear ODE:

y = - x^{4} - 6 x^{2} + c_{1} x + \frac{4 x ^{4}}{3} - 9

y = - x^{4} - 6 x^{2} + c_{1} x + \frac{4 x ^{4}}{3} - 9

n = 1 \sum \infty \frac{n !}{n}

x \to 0 lim (- 7 x^{2} + 7)

\int \frac{z + 5}{z ( z ^{2} + 4 )} d z

\frac{\partial}{\partial v} (u v^{2} + w^{3})

\int e^{5 x + 6} d x

What is the solution for (dy)/(dx)(x)=y+(x^2-3)^2 ?
The solution for (dy)/(dx)(x)=y+(x^2-3)^2 is y=-x^4-6x^2+c_{1}x+(4x^4)/3-9