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#### karush

##### Well-known member

- Jan 31, 2012

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$$\displaystyle y^\prime +2y =xe^{-2x}, \quad y(1)=0$$

$$u(x)=exp\int2 \, dx = e^{2x} $$

$$u(x)=exp\int2 \, dx = e^{2x} $$

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- Thread starter
- #1

- Jan 31, 2012

- 2,928

$$\displaystyle y^\prime +2y =xe^{-2x}, \quad y(1)=0$$

$$u(x)=exp\int2 \, dx = e^{2x} $$

$$u(x)=exp\int2 \, dx = e^{2x} $$

- Mar 10, 2012

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What is the question?$$\displaystyle y^\prime +2y =xe^{-2x}, \quad y(1)=0$$

$$u(x)=exp\int2 \, dx = e^{2x} $$

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Your integrating factor is correct, but I would use the notation:is that correct

if so ill proceed

\(\displaystyle \mu(x)=\exp\left(2\int \,dx\right)=e^{2x}\)

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- Jan 31, 2012

- 2,928

$$u(x)=exp\int2 \, dx = e^{2x} $$

$$\displaystyle y^\prime +2y =xe^{-2x}, \quad y(1)=0$$

obtain u(x)

$$u(x)=exp\int2 \, dx = e^{2x} $$

distribute

$$e^{2x} y'+e^{2x} 2y=e^{2x} xe^{2x}$$

simplify

$$\frac{dy}{dx}(e^{2x} (2y))= 4xe^x $$

not sure

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okYour integrating factor is correct, but I would use the notation:

\(\displaystyle \mu(x)=\exp\left(2\int \,dx\right)=e^{2x}\)

i kinda took some more steps but ...

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- Jan 31, 2012

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$$\displaystyle y^\prime +2y =xe^{-2x}, \quad y(1)=0$$

obtain u(x)

$$u(x)=exp\int2 \, dx = e^{2x} $$

distribute

$$e^{2x} y'+e^{2x} 2y=e^{2x} xe^{-2x}=x$$

simplify

$$ye^{(-2x)}=\int x dx = \frac{x^2}{2}+c$$

divide

$$y=\frac{x^2 e^{2x}}{2}+ce^{2x}$$

then

$$y(1)=\frac{e^{2}}{2}+ce^2=0 \therefore c=-\frac{1}{2}$$

finally

$$y=\frac{1}{2}(x^2-1)e^2$$

so any oops stuff

obtain u(x)

$$u(x)=exp\int2 \, dx = e^{2x} $$

distribute

$$e^{2x} y'+e^{2x} 2y=e^{2x} xe^{-2x}=x$$

simplify

$$ye^{(-2x)}=\int x dx = \frac{x^2}{2}+c$$

divide

$$y=\frac{x^2 e^{2x}}{2}+ce^{2x}$$

then

$$y(1)=\frac{e^{2}}{2}+ce^2=0 \therefore c=-\frac{1}{2}$$

finally

$$y=\frac{1}{2}(x^2-1)e^2$$

so any oops stuff

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