Question

Calculate y as a function of x when dy/dx = 4x3 + 3x2 - 6x + 5

Answer 1

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.

   $\begin{aligned} \dfrac{dy}{dx} &= 4x^3 + 3x^2 - 6x + 5 \\ dy &= (4x^3 + 3x^2 - 6x + 5) dx \\ \int dy &= \int (4x^3 + 3x^2 - 6x + 5) dx \end{aligned}$

In general (understood that +C portions are still there), 

   $\int x^{m} = \dfrac{x^{m+1}}{m+1}$

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

   $\int \big[f(x) \pm g(x)\big]\, dx = \int f(x)\,dx \pm \int g(x) \, dx$

Applying these concepts:

   $\begin{aligned} \int dy &= \int (4x^3 + 3x^2 - 6x + 5) \, dx \\ y &= \int 4x^3\,dx + \int 3x^2 \, dx - \int 6x\, dx + \int 5\, dx \\ &= \frac{4x^4}{4} + \frac{3x^3}{3} - \frac{6x^2}{2} + 5x + C \qquad \text{(only one C is needed)} \\ &= x^4 + x^3 - 3x^2 + 5x + C \end{aligned}$

The answer is y = x⁴ + x³ - 3x² + 5x + C