Differential Equations.

Question

I've been trying to solve this problem for hours.

 

Find the solution of the differential equation

(ln(y))^4  dy/dx=x^4*y

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

y=?

Answer 1

Separate variables:

([ln(y)]^4 / y) dy = x^4 dx

 

U-sub with u=ln(y), du=dy/y:

u^4 du = x^4 dx

 

Integrate:

u^5 = x^5 +C  (integrating will give you coefficients of 1/5, but multiplying the entire equation by 5 cancels them out. it changes the value of C to 5C, but that is still just an arbitrary constant and we can leave it as-is.)

 

[ln(y)]^5 = x^5 +C

 

ln(y) = (x^5 +C) ^(1/5) (make sure the C is inside the 5th root--you cannot simplify the x^5 here)

 

y = e^[(x^5+C)^(1/5)]

 

Initial condition:

e^2 = e^[(1^5+C)^(1/5)]

2=(1+C)^(1/5)

2^5 = 1+C

32=1+C

C=31

 

y = e^[(x^5+31)^(1/5)]