Differential Equations — Maths STD 12 Science — Question

To prove: y given by eq. (i) is a solution of differential equation

(y sin y + cosy + x)y' = y ....(ii)

Proof: Differentiating both sides of eq. (i) w.r.t x, we have

y' + (sin y)y' = 1

$\Rightarrow y'\left( {1 + \sin y} \right) = 1$

$ \Rightarrow y'=\frac{1}{{1 + \sin y}}$ ....(iii)

Putting the value of x from eq. (i) and value of y' from eq. (iii) in L.H.S. of eq. (ii),

(y sin y + cos y + x)y'

$\Rightarrow \left( {y\sin y + \cos y + y - \cos y} \right)\frac{1}{{1 + \sin y}}$

$ \Rightarrow \left( {y\sin y + y} \right)\frac{1}{{1 + \sin y}}$

$ \Rightarrow y\left( {\sin y + 1} \right)\frac{1}{{1 + \sin y}} = y$ = R.H.S. of (ii)

Hence, Function given by eq. (i) is a solution of (y sin y + cos y + x)y' = y.