Question

Solution of differential equation $\sin y.\frac{dy}{dx}+\frac{1}{x}\cos y=x^4{\cos }^{2}y$ is

• A. $x\sec y=x^6+C$
• B. $6x\sec y=x+C$
• C. $6x\mathrm{secy}=x^6+C$
• D. $6x\sec y=6x^6+C$

Answer 1

Answer: (C)

$6x\mathrm{secy}=x^6+C$

$siny\times \frac{dy}{dx}+\frac{1}{x}cosy=x^{4}co{s}^{2}y$

divide both side by $siny$

$=>\frac{dy}{dx}+\frac{1}{x}\frac{cosy}{siny}=\frac{x^{4}co{s}^{2}y}{siny}$

$=>\frac{dy}{dx}+\frac{1}{x}coty=x^{4}cotycosy$

Now this in the form of

$\frac{dy}{dx}+py=q------------\left(i\right)$

$p=\frac{1}{x},q=x^{4}cotycosy$

first we solve Integreted factor

$=>e^{\int }\frac{1}{x}dx$

$=>e^{(}logx)$

$=>x$

now put the value if $p$ and $d$ in equation $\left(i\right)$ we get,

$=>y\times x=\int x^4\times coty\times cosx\times xdx$

$=>xy=\int x^{5}cotycosxdx$

$=>xy=\frac{x^{6}cotycosy}{6}$

$=>6xsecy=x^6+c$