Question #7490

Solve differential equation
-xdy/dx+6y=3x(y^4)^1/3

Expert's answer

First you need to divide every term in the equation by x:

y'+6/x*y=3y^(4/3)

This equation is often called Bernoulli differential equation. You can find here more sufficient information about it and how to solve it.

Then we divide every term of the equation by y^(4/3):

y^(-4/3)*y'+6/x*y^(-1/3)=3

Now we make a substitution z=y^(-4/3+1)=y^(-1/3) => y=1/z^3, y'=-3/z^4 *z'

So we obtain

-3/z^4*z'*z^4+6/x*z=3

-3z'+6z/x=3

z'-2z/x=-1

z(x)=Cx^2-x

So y=1/z^3=1/(Cx^2-x)

y'+6/x*y=3y^(4/3)

This equation is often called Bernoulli differential equation. You can find here more sufficient information about it and how to solve it.

Then we divide every term of the equation by y^(4/3):

y^(-4/3)*y'+6/x*y^(-1/3)=3

Now we make a substitution z=y^(-4/3+1)=y^(-1/3) => y=1/z^3, y'=-3/z^4 *z'

So we obtain

-3/z^4*z'*z^4+6/x*z=3

-3z'+6z/x=3

z'-2z/x=-1

z(x)=Cx^2-x

So y=1/z^3=1/(Cx^2-x)

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