Answer:
y = x⁴ + x³ - 3x² + 5x + C
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Separable differential equations such as these ones can be solved by treating dy/dx as a ratio of differentials. Then move the dx with all the x terms and move the dy with all the y terms. After that, integrate both sides of the equation.
[tex]\begin{aligned}
\dfrac{dy}{dx} &= 4x^3 + 3x^2 - 6x + 5 \\
dy &= (4x^3 + 3x^2 - 6x + 5) dx \\
\int dy &= \int (4x^3 + 3x^2 - 6x + 5) dx
\end{aligned}[/tex]
In general (understood that +C portions are still there),
[tex]\int x^{m} = \dfrac{x^{m+1}}{m+1}[/tex]
Note that ∫dy = y since it is ∫1·dy = ∫y⁰ dy = y¹/(0+1) = y
For the right-hand side, we use the sum/difference rule for integrals, which says that
[tex]\int \big[f(x) \pm g(x)\big]\, dx = \int f(x)\,dx \pm \int g(x) \, dx[/tex]
Applying these concepts:
[tex]\begin{aligned}
\int dy &= \int (4x^3 + 3x^2 - 6x + 5) \, dx \\
y &= \int 4x^3\,dx + \int 3x^2 \, dx - \int 6x\, dx + \int 5\, dx \\
&= \frac{4x^4}{4} + \frac{3x^3}{3} - \frac{6x^2}{2} + 5x + C \qquad \text{(only one $C$ is needed)} \\
&= x^4 + x^3 - 3x^2 + 5x + C
\end{aligned}[/tex]
The answer is y = x⁴ + x³ - 3x² + 5x + C
