Differential Equations

Question

Find the solution of the differential equation

 

dy/dx*(ln(y))^2 = yx^2

 

which satisfies the initial condition y(1)=e^2

 

y = ?

Answer 1

Hi Ally, this is a separable equation. First step to take is to rearrange the equation so that you obtain:

 

f (y)dy = g (x)dx

 

We want to group y terms with dy and x terms with dx, doing so yields:

 

(ln (y))^2/y *dy = x^2dx

Now you find the antiderivative of each side:

 

The right side antiderivative is:  x^3/3 + C

 

The left side antiderivative requires a substitution.

 

Let u = lny 

This is a good voice since the derivative of lny is 1/y, and is present in the left side.

 

Differentiate w respect to y: du/dy= 1/y

 

Rearranging, du = dy/y 

 

Substituting u = lny and du = dy/y into the left hand side yields: u^2 du

 

Antidifferentiating yields u^3/3

 

Replacing u with ln (y) converts the antiderivative to the original variable y.

 

(ln (y))^3/3  is the left side antiderivative.

 

Equating the two antiderivatives yields:

 

(ln (y))^3/3 = x^3/3 + c 

Multiply by 3 and exponentiate to obtain:

 

Y (x) = e^(x^3 + 3c)

Apply y (1) = e^2 to obtain

 

Y (1) = e^(1^3+ 3c) So...

e^2 = e^(1+3c),  since the bases are equal we can equate the exponents to obtain

 

2= 1+ 3c, therefore c = 1/3

Substituting c = 1/3 into Y (x) yields

 

Y (x) = e^[x^3+ 3 (1/3)]

 

Y (x) = e^[x^3 + 1] , final solution

 

I hope this helps!

 

Wyatt