Question

solve the linear differential equation (1+x)dy/dx-xy=1-x

Answer 1

i would suggest plzz repost the ques..

Answer 2

Answer:

${ e }^{ -x }(1+x)y=(x-1){ e }^{ -x }+{ e }^{ -x }+C$

Step-by-step explanation:

$(1+x)\frac { dy }{ dx } -xy=1-x\\ \frac { dy }{ dx } -\frac { x }{ 1+x } y=\frac { 1-x }{ 1+x } \\ it\quad is\quad in\quad the\quad form\quad of\quad \frac { dy }{ dx } +p\left( x \right) =q\left( x \right)$

$Here\quad p\left( x \right) =\frac { -x }{ 1+x } ,\quad q\left( x \right) =\frac { 1-x }{ 1+x } \\ Integration\quad Factor\quad (IF)={ e }^{ \int { p\left( x \right) dx }  }\\ \qquad \qquad \qquad \qquad \qquad \qquad ={ e }^{ \int { \frac { -x }{ 1+x } dx }  }={ e }^{ -x }(1+x)$

$General\quad Solution\quad \\ (IF)y=\int { q\left( x \right) (IF)dx } \\ { e }^{ -x }(1+x)y=\int { { e }^{ -x }(1+x) } \frac { 1-x }{ 1+x } dx\\ { e }^{ -x }(1+x)y=\int { { e }^{ -x } } (1-x)dx\\ { e }^{ -x }(1+x)y=(x-1){ e }^{ -x }+{ e }^{ -x }+C$