How do you solve the differential #dy/dx=(10x^2)/sqrt(1+x^3)#?

Question

Aurora Alvarado · Accepted Answer

#y = 20/3 sqrt(1+x^3)+C# We know that #d/dx(sqrt(1+x^3))=3/2 x^2/sqrt(1+x^3)# so, grouping variables #dy =20/3d/dx(sqrt(1+x^3))dx # integrating #y = 20/3 sqrt(1+x^3)+C#

Isabella Ackerman · Answer

undefined To solve the differential equation $ \frac{dy}{dx} = \frac{10x^2}{\sqrt{1+x^3}} $, you can separate variables and integrate both sides. Here's the step-by-step process:

1. Separate variables:
   $ \frac{dy}{\sqrt{1+y^3}} = 10x^2 \, dx $

2. Integrate both sides:
   $ \int \frac{1}{\sqrt{1+y^3}} \, dy = \int 10x^2 \, dx $

3. For the left integral, you can use the substitution $ u = 1+y^3 $, so $ du = 3y^2 \, dy $:
   $ \frac{1}{3} \int \frac{1}{\sqrt{u}} \, du = \int 10x^2 \, dx $

4. After integrating both sides:
   $ \frac{2}{3} \sqrt{u} = \frac{10}{3} x^3 + C $

5. Substitute $ u = 1+y^3 $ back in:
   $ \frac{2}{3} \sqrt{1+y^3} = \frac{10}{3} x^3 + C $

6. Solve for $ y $:
   $ \sqrt{1+y^3} = 5x^3 + C' $

7. Square both sides to isolate $ y $:
   $ 1+y^3 = (5x^3 + C')^2 $

8. Subtract 1 from both sides:
   $ y^3 = (5x^3 + C')^2 - 1 $

9. Finally, solve for $ y $ by taking the cube root:
   $ y = \sqrt[3]{(5x^3 + C')^2 - 1} $