Answer:
[tex]\boxed{y = -\dfrac{2}{289x^2 + C}}[/tex]
Step-by-step explanation:
First, we can rewrite y' with Leibniz notation:
[tex]\dfrac{dy}{dx} = 289xy^2[/tex]
Next, we want to separate y's and x's onto different sides. We can do this by:
↓↓↓
[tex]\dfrac{1}{y^2}\ dy= 289x\ dx[/tex]
Now, we can take the indefinite integral (or antiderivative) of both sides:
[tex]\displaystyle \int y^{-2}\ dy= \int 289x\ dx[/tex]
↓ taking out the constant on the right side
[tex]\displaystyle \int y^{-2}\ dy= 289\!\left(\int x\ dx\right)[/tex]
↓ evaluating the integrals using the power rule: [tex]\int x^n \ dx = \left(\frac{1}{n+1}\right)x^{n+1}+C[/tex]
[tex]\dfrac{1}{-2+1}\,y^{(-2+1)} = 289\left(\dfrac{1}{1+1}\,x^{1+1}\right) + C[/tex]
↓ simplifying the fractions and exponents
[tex]-\dfrac{1}{1} \cdot \dfrac{1}{y} = 289\cdot \dfrac{1}{2} \,x^2 + C[/tex]
↓ multiplying both sides by -1
[tex]\dfrac{1}{y} = -\dfrac{289x^2 + C}{2}[/tex]
↓ taking the reciprocal of both sides
[tex]\boxed{y = -\dfrac{2}{289x^2 + C}}[/tex]
