Welcome to our community

Be a part of something great, join today!

[SOLVED] 2.1.6 de with IVP

karush

Well-known member
Jan 31, 2012
2,928
$$\displaystyle y^\prime +2y =xe^{-2x}, \quad y(1)=0$$
$$u(x)=exp\int2 \, dx = e^{2x} $$
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
835

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
is that correct

if so ill proceed
Your integrating factor is correct, but I would use the notation:

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

karush

Well-known member
Jan 31, 2012
2,928
$$\displaystyle y^\prime +2y =xe^{-2x}, \quad y(1)=0$$
$$u(x)=exp\int2 \, dx = e^{2x} $$

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

- - - Updated - - -

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

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

i kinda took some more steps but ...
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I think what you mean is:

\(\displaystyle \frac{d}{dx}\left(e^{2x}(2y)\right)\)

Please perform the indicated differentiation (product rule), and see if you get back to the previous step. :)
 

karush

Well-known member
Jan 31, 2012
2,928
$$\displaystyle y^\prime +2y =xe^{-2x}, \quad y(1)=0$$
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
 
Last edited: