Dif-eq particular solution to xy'ln(x) + y = x² ln(x)

[scaryhans]

Dif-eq particular solution to xy'ln(x) + y = x² ln(x)

I have to find the general solution of the following equation: xy'ln(x) + y = x² ln(x)
i have arranged this differently to fit : y' + f(x)y = g(x)
=> y' + y/(xln(x)) = x
for this equation i have found the following solution for the complementary equation:
=> y' + y/(xln(x)) = 0 => y = -ln(x)*C
now i need to figure out the particular solution, for which i have the following g(x) = x
=> Yp = Ax + B => Y'p = A
in the original equation:
=> y' + y/(xln(x)) = x => A + (Ax + B)/(xln(x)) = x
and this is where i am a bit stuck, i don't know how to continu with this particular equation to eventually get to the general solution of this differential equation. Any help is welcome!

 

[Deleted member 4993]

I have to find the general solution of the following equation: xy'ln(x) + y = x² ln(x)
i have arranged this differently to fit : y' + f(x)y = g(x)
=> y' + y/(xln(x)) = x
for this equation i have found the following solution for the complementary equation:
=> y' + y/(xln(x)) = 0 => y = -ln(x)*C
now i need to figure out the particular solution, for which i have the following g(x) = x
=> Yp = Ax + B => Y'p = A
in the original equation:
=> y' + y/(xln(x)) = x => A + (Ax + B)/(xln(x)) = x
and this is where i am a bit stuck, i don't know how to continu with this particular equation to eventually get to the general solution of this differential equation. Any help is welcome!

This is a first order equation - do not need to go homogeneous/particular solution route:

y' + f(x) * y = g(x)

\(\displaystyle y' * e^{\int f(x)dx} + f(x) * y * e^{\int f(x)dx} \ = \ g(x)* e^{\int f(x)dx}\)

\(\displaystyle \displaystyle{y * e^{\int f(x)dx} \ = \ \int \left [g(x)* e^{\int f(x)dx}\right ] dx}\)

and continue...

 

[scaryhans]

This is a first order equation - do not need to go homogeneous/particular solution route:

y' + f(x) * y = g(x)

\(\displaystyle y' * e^{\int f(x)dx} + f(x) * y * e^{\int f(x)dx} \ = \ g(x)* e^{\int f(x)dx}\)

\(\displaystyle \displaystyle{y * e^{\int f(x)dx} \ = \ \int \left [g(x)* e^{\int f(x)dx}\right ] dx}\)

and continue...

I hadn't thought of that. But i have found an error in my homogenous solution it should be C/ln(x). I applied variation of parameters on this, with C(x)/ln(x)

From this i found that C'(x)=x*ln(x) => C(x) = [x²(2ln(x)-1)]/4 + k

and this gives us our general solution: y = [x²(2ln(x)-1)]/4ln(x) + k/ln(x)
does this seem right?

 

[Deleted member 4993]

I hadn't thought of that. But i have found an error in my homogenous solution it should be C/ln(x). I applied variation of parameters on this, with C(x)/ln(x)

From this i found that C'(x)=x*ln(x) => C(x) = [x²(2ln(x)-1)]/4 + k

and this gives us our general solution: y = [x²(2ln(x)-1)]/4ln(x) + k/ln(x)
does this seem right?

You can always check the solution of ODE by differentiating your solution and put it back in the ODE and see if you it satisfies that equation!

 

[Deleted member 4993]

This is a first order equation - do not need to go homogeneous/particular solution route:

y' + f(x) * y = g(x)

\(\displaystyle y' * e^{\int f(x)dx} + f(x) * y * e^{\int f(x)dx} \ = \ g(x)* e^{\int f(x)dx}\)

\(\displaystyle \displaystyle{y * e^{\int f(x)dx} \ = \ \int \left [g(x)* e^{\int f(x)dx}\right ] dx}\)

and continue...

\(\displaystyle \displaystyle{y * e^{ln[ln(x)]} \ = \ \int \left [x* e^{ln[ln(x)]}\right ] dx}\)

\(\displaystyle \displaystyle{y * ln(x) \ = \ \int x* ln(x) dx}\)

\(\displaystyle \displaystyle{y * ln(x) \ = \ \frac{x^2}{2} * \left [ln(x) -\frac{1}{2}\right ] \ + \ C}\)