Question

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

A. x4 + x3 - 3x2 + 5x + C
B. 12x2 +3x + C
C. 4x4/3 + x3 - 6x2 + 5x + C
D. 4x2 +3x + C

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} (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) = (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 \\ &= (4x^4)/(4) + (3x^3)/(3) - (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