Question

FIND IT................↑↑

Answer 1

$xdy/dx + 2y = x {}^{2} log(x) \\ divide \: b.s \: by \: x \\ dy/dx + 2y/x = x log(x) \\ comparing \: with \: dy/dx + py = q \\ where.p = 2/x.q = x log(x) \\ i.f =e {}^{∫pdx} \\ = e {}^{∫2/xdx} \\ = e {}^{2∫1/xdx} \\ = e {}^{2 log(x) } \\ = e {}^{ log(x) {}^{2} } \\ = x {}^{2} \\ so \: the \: \: solution \: is \\ y(i.f) = ∫q(j.f)dx + c \\ y(x {}^{2} ) =∫(x log(x) ) x {}^{2} dx + c \\ x {}^{2} y = ∫x {}^{3} log(x) dx + c \\ x {}^{2}y =x {}^{4}/4( log(x) ) -∫(1/x)x {}^{4}/4 \\ (since \: ∫udv = uv-∫vdu) \\ x {}^{2} y = x {}^{4}/4( log(x))-∫x {}^{3}/4 \\ x {}^{2} y = x {}^{4}/4log(x) - 1/4∫x {}^{3} \\ x {}^{2} y = x {}^{4}/4 log(x) -1/4(x {}^{4}/4) + c \\ x {}^{2} y = x {}^{4}/4(( log(x) - 1/4) + c \\ y = x {}^{2}/4( log(x -1/4 ) + c \\ y = x {}^{2} /4( log(x) ) - x {}^{2}/16 + c$

Where 'c' is constant

Answer 2

$\huge\bf\red{AnswEr:-}$

$$\begin{lgathered}xdy/dx + 2y = x {}^{2} log(x) \\ divide \: b.s \: by \: x \\ dy/dx + 2y/x = x log(x) \\ comparing \: with \: dy/dx + py = q \\ where.p = 2/x.q = x log(x) \\ i.f =e {}^{∫pdx} \\ = e {}^{∫2/xdx} \\ = e {}^{2∫1/xdx} \\ = e {}^{2 log(x) } \\ = e {}^{ log(x) {}^{2} } \\ = x {}^{2} \\ so \: the \: \: solution \: is \\ y(i.f) = ∫q(j.f)dx + c \\ y(x {}^{2} ) =∫(x log(x) ) x {}^{2} dx + c \\ x {}^{2} y = ∫x {}^{3} log(x) dx + c \\ x {}^{2}y =x {}^{4}/4( log(x) ) -∫(1/x)x {}^{4}/4 \\ (since \: ∫udv = uv-∫vdu) \\ x {}^{2} y = x {}^{4}/4( log(x))-∫x {}^{3}/4 \\ x {}^{2} y = x {}^{4}/4log(x) - 1/4∫x {}^{3} \\ x {}^{2} y = x {}^{4}/4 log(x) -1/4(x {}^{4}/4) + c \\ x {}^{2} y = x {}^{4}/4(( log(x) - 1/4) + c \\ y = x {}^{2}/4( log(x -1/4 ) + c \\ y = x {}^{2} /4( log(x) ) - x {}^{2}/16 + c\end{lgathered}$$

• Where 'c' is constant✔✔