Hi everyone! My question is: "find the general solution y(x) of the given differential equation": #xy'+y=x^6# How do I solve this? Thank you for your answer!

Here:
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This should be answer, but I'm not sure how to get to that exact point:
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2 Answers
Andrea S.
May 17, 2018

#y(x) = 1/(7x) (c+x^7)#

Explanation:

Solve the homogeneous equation:

#xy'+y = 0#

which is separable:

#x dy/dx = -y#

#dy/y = -dx/x#

#ln y = -ln x +c#

#y=c/x#

Using now the variable coefficients methods, look for a solution of the complete equation in the form:

#bar y = (c(x))/x#

substitute #bary# in the complete equation:

#xd/dx( (c(x))/x ) + (c(x))/x = x^6#

#x (xc'(x)-(c(x)))/x^2 + (c(x))/x = x^6#

#c'(x) - (c(x))/x+ (c(x))/x = x^6#

#c'(x) = x^6#

#c(x) = int x^6dx = x^7/7+c_1#

Take the solution with #c_1=0#

#bary = (c(x))/x = x^6/7#

Then the general solution of the equation is:

#y(x) =c/x + x^6/7 = 1/(7x) (c+x^7)#

Steve M
May 18, 2018

# y = 1/(7x)(x^7 + C) #

Explanation:

We have:

# xy'+y=x^6 #

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

# dy/dx + P(x)y=Q(x) #

So, we can put the equation in standard form:

# y'+1/x y = x^5 #

Then the integrating factor is given by;

# I = e^(int P(x) dx) #
# \ \ = exp(int \ 1/x \ dx) #
# \ \ = exp(lnx) #
# \ \ = x #

And if we multiply the DE by this Integrating Factor, #I#, we will have a perfect product differential (in fact the original equation);

# xy'+y=x^6 #

# :. d/dx (xy) = x^6 #

This is now separable, so by "separating the variables" we get:

# xy = int \ x^6 \ dx #

Which is trivial to integrate to get

# xy = x^7/7 + C #

Leading to the GS:

# y = 1/(7x)(x^7 + C) #

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