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Given:
The equation of line is
To find:
The x-intercept and y-intercept.
Solution:
The equation of line is
Putting x=0, we get
So, the y-intercept is (0,3).
Putting y=0 in the given equation, we get
So, the y-intercept is (-3,0).
Therefore, the coordinates of the point where the line intersects the y-axis are (0,3). The coordinates of the point where the line intersects the x-axis are (-3,0).
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.
In general (understood that +C portions are still there),
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](https://tex.z-dn.net/?f=%5Cint%20%5Cbig%5Bf%28x%29%20%5Cpm%20g%28x%29%5Cbig%5D%5C%2C%20dx%20%3D%20%5Cint%20f%28x%29%5C%2Cdx%20%5Cpm%20%5Cint%20g%28x%29%20%5C%2C%20dx)
Applying these concepts:
The answer is y = x⁴ + x³ - 3x² + 5x + C
y = x⁴ + x³ - 3x² + 5x + C
======
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.
In general (understood that +C portions are still there),
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
Applying these concepts:
The answer is y = x⁴ + x³ - 3x² + 5x + C