What is the general solution of the differential equation

#dy/dx+x^4y=3x^4 (x0)#?

Express the answer in explicit form.

Question

I understand that the equation is of the form that uses the integrating factor method:

#dy/dx+g(x)y=h(x)#,

with #g(x)=x^4#, and #h(x)=3x^4#.

The integrating factor

#p(x)=exp(intx^4dx)=exp((x^5)/5)=e^((x^5)/5)#

The general solution is

#y=1/(p(x))(intp(x)h(x) dx)#

#=1/e^((x^5)/5)(int3x^4e^((x^5)/5)dx)#

..... and going on from here is where I get confused... very basic step by step explanation from here on would be very gratefully received.

Evelyn Agard · Accepted Answer

The solution is #y=3-C_1e^(-x^5/5)#

I see a simpler solution.

#dy/dx+x^4y=3x^4#

#dy/dx=3x^4-x^4y#

#dy/dx=x^4(3-y)#

We separate the variables

#dy/(3-y)=x^4dx#

Integrating both sides

#intdy/(3-y)=intx^4dx#

#-ln(3-y)=x^5/5+C#

#ln(3-y)=-x^5/5-C#

#(3-y)=e^(-x^5/5-C)#

#3-y=C_1e^(-x^5/5)#

#y=3-C_1e^(-x^5/5)#

Paisley Johnson · Answer

The answer is #y=3+Ce^(-x^5/5)#

We can apply your method

The integrating factor is #e^(x^5/5)#

#dy/dx+yx^4=3x^4#

Multiplying by #e^(x^5/5)#

#e^(x^5/5)dy/dx+yx^4e^(x^5/5)=3x^4e^(x^5/5)#

#d/dx(ye^(x^5/5))=3x^4e^(x^5/5)#

Integrating both sides

#intd/dx(ye^(x^5/5))=int3x^4e^(x^5/5)dx#

#ye(x^5/5))=3intx^4e^(x^5/5)dx#

For the #RHS#,

Let #u=x^5/5#, #=>#, #du=x^4dx#

#intx^4e^(x^5/5)dx=inte^(u)du#

#=e^(u)=e^(x^5/5)+C#

Therefore,

#ye(x^5/5))=3e^(x^5/5)+C#

#y=3+Ce^(-x^5/5)#