Question

how to solve this differential equation

Answer 1

\[e^{x}\frac{ dy }{ dx } +1 = x \]

Answer 2

\[e^x \frac{ dy }{ dx }+1=x\]
\[e^x \frac{ dy }{ dx }=x-1\]
\[dy=\left( x+1 \right)e ^{-x}dx\]
integrate it.

Answer 3

correction it is x-1 in the end.

Answer 4

@surjithayer i knew how to rearrange this (that is most basic) .. i am not getting the integration

Answer 5

\[y=\int\limits \left( x-1 \right)e ^{-x}dx+c\]
integrate by parts.

Answer 6

\[- e^{-x} (x-1) - \int\limits e^{-x} dx\]

Answer 7

is this correct ? @surjithayer

Answer 8

\[- e^{-x} (x-1) + e^{-x} +c \]

Answer 9

but this answer given is \[y= -x e^{-x} +c \]

Answer 10

\[y=-\left( x-1 \right)e ^{-x}-\int\limits 1.\left( -e ^{-x} \right)dx+c\]
\[y=-\left( x-1 \right)e ^{-x}+\int\limits e ^{-x}dx+c\]
\[y=-x e ^{-x}+e ^{-x}-e ^{-x}+c\]
?

Answer 11

why have you written \[y=-\left( x-1 \right)e ^{-x}-\int\limits\limits 1.\left( -e ^{-x} \right)dx+c\] -e^(-x) dx ??

Answer 12

what does the integration by parts rule say ?? @surjithayer my question is why the minus sign ahead of e^(-x) after the integral sign

Answer 13

\[\int\limits uv dx=u \int\limits v dx-\int\limits \frac{ du }{ dx }*(\int\limits v dx)dx\]

Answer 14

\[\int\limits (1st)(2nd) dx=1st \int\limits~ 2nd~ dx-\int\limits ~(differential~of~1st)*(\int\limits 2nd~dx)~dx\]

Answer 15

\[\int\limits e ^{-x}dx=\frac{ e ^{-x} }{ -1 }=- e ^{-x}\]

Answer 16

ok got it @surjithayer ...thanks :)

Answer 17

yw