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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS3 Marks
Question
In each of the verify that the given function (explicit or implicit) is a solution of the corresponding differential equation: $\text{y} – \text{cos}\ \text{y} = \text{x} \ :\ (\text{y} \ \text{sin} \ \text{y} + \text{cos} \ \text{y} + \text{x}) \text{y}' = \text{y}$
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Answer

Given: y-cos y = x ....(i)
To prove: y given by eq. (i) is a solution of differential equation
(y sin y + cos y + x) y' = y ....(ii)
Proof: Differentiating both sides of eq. (i) w.r.t x, we have

$\text{y}'+(\text{sin}\ \text{y})\text{y}'=1 \ \Rightarrow\ \ \text{y}'(1+\text{sin}\ \text{y})=1$
$\Rightarrow\ \ \text{y}'=\frac{1}{1+\text{sin}\ \text{y}} \ ....(\text{iii})$
Putting the value of x from eq. (i) and value of y' from eq. (iii) in L.H.S. of eq. (ii),

$\text{(y sin y + cos y + x)}{\text{y}}' $ $\ \Rightarrow \ \text{(y sin y + cos y + y} -\text{cos} \ \text{y}) \frac{1}{1+\text{sin}\ \text{y}}$
$\Rightarrow \ \text{(y sin y + y)}\frac{1}{1+ \text{sin y}} $ $\ \Rightarrow \ \text{y}(\text{sin y} + 1)\frac{1}{1+\text{sin y}}= \text{y = R.H.S. of (ii)}$
Hence, Function given by eq. (i) is a solution of (y sin y + cos y + x) y' = y.

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