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

Answer 1

#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#
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Answer 2

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} )

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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