Question

HELP PLEASE. My Last Problem of Homework. :/

Answer 1

@518nad

Answer 2

Sorry for the messy handwriting, but here is my work. Its #10

Answer 3

Would save you grief if you wrote as

\((x^2 +1)u' + 3x u = 0\)

\(u = y - 1, u' = y'\)

keeps it homogeneous all the way

then use an integrating factor

Answer 4

u made mistake with multiplying integrating factor wit the right side check that over

Answer 5

i do like that simplification

Answer 6

thus u get
u=e^int(-3x/(x^2+1) dx )

y= e^int(-3x/(x^2+1) dx ) +1

Answer 7

\[y = e^{ \left( {\int\limits \frac{ -3x }{ x^2+1 }dx}\ \right) +1 }\] How did you get that?

Answer 8

+1 is not exponent

Answer 9

\[\left( x^2+1 \right)\frac{ dy }{ dx }+3xy=3x\]
\[divide~ by~ (x^2+1)\]
\[\frac{ dy }{ dx }+\frac{ 3x }{ x^2+1 }y=\frac{ 3x }{ x^2+1 }\]
\[I.F.=e ^{\int\limits \frac{ 3x }{ x^2+1 }dx}=e ^{\frac{ 3 }{ 2 } \int\limits \frac{ 2x }{ x^2+1 }dx }=e ^{\frac{ 3 }{ 2 }\ln \left( x^2+1 \right)}\]
\[=e ^{\ln \left( x^2+1 \right)^{\frac{ 3 }{ 2 }}}=\left( x^2+1 \right)^{\frac{ 3 }{ 2 }}\]
c.s. is
\[y \left( x^2+1 \right)^{\frac{ 3 }{ 2 }}=\int\limits \frac{ 3x }{ x^2+1 }\left( x^2+1 \right)^{\frac{ 3 }{ 2 }}dx+c\]
\[=\frac{ 3 }{ 2 }\int\limits \left( x^2+1 \right)^{\frac{ 1 }{ 2 }}2x dx+c\]
i think you can complete now.

Answer 10

Why did you add a constant before you even finished integrating?

Answer 11

@mathmale @pooja195

Answer 12

it makes no difference if you add constant here or at the end.

Answer 13

my aim was to tell ,you have to add a constant.