Diff Equation Problem - # 3

Jason76

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Show that \(\displaystyle y = x(\ln(x))^{2} + Cx\) is a general solution to

\(\displaystyle xy' - y = 2x\ln(x)\)

\(\displaystyle y' - \dfrac{y}{x} = \dfrac{2x\ln(x)}{x}\)

\(\displaystyle e^{\int -\dfrac{1}{x} (dx)} = e^{-\ln(x)} = e^{\ln(x^{-1})} = x^{-1} = \dfrac{1}{x}\)

\(\displaystyle y'\dfrac{1}{x} - \dfrac{y}{x}\dfrac{1}{x} = \dfrac{2x\ln(x)}{x}\dfrac{1}{x}\)

\(\displaystyle \dfrac{d}{dx}[\dfrac{1}{x}y ] = \dfrac{2x\ln(x)}{x^{2}}\)

\(\displaystyle \int \dfrac{d}{dx}[\dfrac{1}{x}y ] (dx) = \int \dfrac{2x\ln(x)}{x^{2}} (dx)\)

\(\displaystyle \int \dfrac{d}{dx}[\dfrac{1}{x}y ] (dx) = \int 2\ln(x)(x^{-1}) (dx)\)

Using integration by parts

\(\displaystyle u = 2 \ln x\)

\(\displaystyle du = \dfrac{2}{x}\)

\(\displaystyle dv = x^{-1}\)

\(\displaystyle v = \ln x\)

\(\displaystyle \int \dfrac{d}{dx}[\dfrac{1}{x}y ] (dx) = 2\ln(x)\ln(x) - \int \ln(x)\dfrac{2}{x}\)


\(\displaystyle \int \dfrac{d}{dx}[\dfrac{1}{x}y ] (dx) = 2\ln(x)\ln(x) - \int 2\ln(x)\dfrac{1}{x}\)

\(\displaystyle xy = 2\ln(x)\ln(x) - 2\ln(x) + C\) ???
 
Here is the possible answer:

Show that \(\displaystyle y = x(\ln(x))^{2} + Cx\) is a general solution to

\(\displaystyle xy' - y = 2x\ln(x)\)

\(\displaystyle y' - \dfrac{y}{x} = \dfrac{2x\ln(x)}{x}\)

\(\displaystyle e^{\int -\dfrac{1}{x} (dx)} = e^{-\ln(x)} = e^{\ln(x^{-1})} = x^{-1} = \dfrac{1}{x}\)

\(\displaystyle y'\dfrac{1}{x} - \dfrac{y}{x}\dfrac{1}{x} = \dfrac{2x\ln(x)}{x}\dfrac{1}{x}\)

\(\displaystyle \dfrac{d}{dx}[\dfrac{1}{x}y ] = \dfrac{2x\ln(x)}{x^{2}}\)

\(\displaystyle \int \dfrac{d}{dx}[\dfrac{1}{x}y ] (dx) = \int \dfrac{2x\ln(x)}{x^{2}} (dx)\)

\(\displaystyle \int \dfrac{d}{dx}[\dfrac{1}{x}y ] (dx) = \int \dfrac{2 \ln(x)}{x} (dx)\)

\(\displaystyle u = \ln(x)\)

\(\displaystyle du = \dfrac{1}{x} dx\)

\(\displaystyle (2)du = (2)\dfrac{1}{x} dx\)

\(\displaystyle \int \dfrac{d}{dx}[\dfrac{1}{x}y ] (dx) = \int 2(u)(du)\)

\(\displaystyle \dfrac{y}{x} = 2\ln(x) + C\)

\(\displaystyle \dfrac{y}{x} = (\ln(x))^{2} + C\) Rewritten using \(\displaystyle \ln\) rules

\(\displaystyle y = x(\ln(x))^{2} + Cx\) This looks right, but was it the right way to find it?
 
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Here is the possible answer:

Show that \(\displaystyle y = x(\ln(x))^{2} + Cx\) is a general solution to

\(\displaystyle xy' - y = 2x\ln(x)\)

\(\displaystyle y' - \dfrac{y}{x} = \dfrac{2x\ln(x)}{x}\)

\(\displaystyle e^{\int -\dfrac{1}{x} (dx)} = e^{-\ln(x)} = e^{\ln(x^{-1})} = x^{-1} = \dfrac{1}{x}\)

\(\displaystyle y'\dfrac{1}{x} - \dfrac{y}{x}\dfrac{1}{x} = \dfrac{2x\ln(x)}{x}\dfrac{1}{x}\)

\(\displaystyle \dfrac{d}{dx}[\dfrac{1}{x}y ] = \dfrac{2x\ln(x)}{x^{2}}\)

\(\displaystyle \int \dfrac{d}{dx}[\dfrac{1}{x}y ] (dx) = \int \dfrac{2x\ln(x)}{x^{2}} (dx)\)

\(\displaystyle \int \dfrac{d}{dx}[\dfrac{1}{x}y ] (dx) = \int \dfrac{2 \ln(x)}{x} (dx)\)

\(\displaystyle u = \ln(x)\)

\(\displaystyle du = \dfrac{1}{x} dx\)

\(\displaystyle (2)du = (2)\dfrac{1}{x} dx\)

\(\displaystyle \int \dfrac{d}{dx}[\dfrac{1}{x}y ] (dx) = \int 2(u)(du)\)

\(\displaystyle \dfrac{y}{x} = u^2 + C\)

\(\displaystyle \dfrac{y}{x} = (\ln(x))^{2} + C\) Rewritten using substitution u = ln(x)

\(\displaystyle y = x(\ln(x))^{2} + Cx\) This looks right, but was it the right way to find it?

Although your answer was correct, your algebra was incorrect.

2*ln(x) \(\displaystyle \ne\) [ln(x)]^2

\(\displaystyle 2*ln(x) = ln(x^2) \)
 
Although your answer was correct, your algebra was incorrect.

2*ln(x) \(\displaystyle \ne\) [ln(x)]^2

\(\displaystyle 2*ln(x) = ln(x^2) \)

I see the \(\displaystyle u^{2}\) comes after doing the power rule and canceling out the 2s.

One big mistake. :!:

Here is correct way:

Show that \(\displaystyle y = x(\ln(x))^{2} + Cx\) is a general solution to

\(\displaystyle xy' - y = 2x\ln(x)\)

\(\displaystyle y' - \dfrac{y}{x} = \dfrac{2x\ln(x)}{x}\)

\(\displaystyle e^{\int -\dfrac{1}{x} (dx)} = e^{-\ln(x)} = e^{\ln(x^{-1})} = x^{-1} = \dfrac{1}{x}\)

\(\displaystyle y'\dfrac{1}{x} - \dfrac{y}{x}\dfrac{1}{x} = \dfrac{2x\ln(x)}{x}\dfrac{1}{x}\)

\(\displaystyle \dfrac{d}{dx}[\dfrac{1}{x}y ] = \dfrac{2x\ln(x)}{x^{2}}\)

\(\displaystyle \int \dfrac{d}{dx}[\dfrac{1}{x}y ] (dx) = \int \dfrac{2x\ln(x)}{x^{2}} (dx)\)

\(\displaystyle \int \dfrac{d}{dx}[\dfrac{1}{x}y ] (dx) = \int \dfrac{2 \ln(x)}{x} (dx)\)

\(\displaystyle u = \ln(x)\)

\(\displaystyle du = \dfrac{1}{x} dx\)

\(\displaystyle (2)du = (2)\dfrac{1}{x} dx\)

\(\displaystyle \int \dfrac{d}{dx}[\dfrac{1}{x}y ] (dx) = \int 2(u)(du)\)

\(\displaystyle \dfrac{y}{x} = 2\dfrac{u^{2}}{2}+ C\) - Needs the power rule :!:


\(\displaystyle \dfrac{y}{x} = u^{2} + C\)

\(\displaystyle \dfrac{y}{x} = (\ln(x))^{2} + C\)

\(\displaystyle y = x(\ln(x))^{2} + Cx\)
 
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This problem does NOT ask you to solve the differential equation! It only asks that you show that \(\displaystyle y(x)= x(ln(x))^2+ Cx\) satisfies \(\displaystyle xy'- y= 2x ln(x)\). That is a much easier problem! If you were asked to "show that x= 2 is a solution to the equation \(\displaystyle x^4- 3x^3- 15x^2+ 99x- 130= 0\)" would you start trying to solve that equation or would you just set x= 2 and evaluate?

If \(\displaystyle y= x(ln(x))^2+ Cx\) then \(\displaystyle y'= 1\cdot (ln(x))^2+ x(2(ln(x))(1/x)+ C= (ln(x))^2+ 2 ln(x)+ C.\)

So

\(\displaystyle xy'- y= x((ln(x))^2+ 2ln(x)+ C)- (x(ln(x))^2+ Cx)\)

\(\displaystyle = x(ln(x))^2+ 2xln(x)+ Cx- x(ln(x))^2- Cx \)

\(\displaystyle = 2xln(x)\)

and we are done!
 
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