Question

Determine the equation of the curve passing through the origin in the form $y=f\left(x\right)$, which satisfies the differential equation $\frac{dy}{dx}=\sin (10x+6y)$.

Answer 1

per of fuels opented engins of dternative verticles

$\frac{dy}{dx}=sin(10x+6y)$

Let $10x+6y=u.$

$10+6\frac{dy}{dx}=\frac{du}{dx}$

$\Rightarrow \frac{1}{6}[\frac{du}{dx}-10]=sinu$

$\frac{du}{dx}=6sinu+10$

$\int \frac{du}{6sinu+10}=\int dx$

$\int \frac{dx}{12sin\frac{u}{2}cos\frac{u}{2}+10}=\int dx$

$\int \frac{se{c}^{2}u/2du}{12tan\frac{x}{2}+10se{c}^2\frac{u}{2}}=\int dx=x+c$

let $tan\frac{x}{2}=t\Rightarrow \frac{1}{2}se{c}^2\frac{x}{2}dx=dt$

$=\int \frac{2dt}{12t+10(1+t^2)}=x+c$

$=\frac{1}{5}\int \frac{dt}{(t+\frac{3}{5}{)}^2+\frac{t^2}{5}}=x+c$

$\frac{1}{5}ta{n}^{-1}\left(\frac{t+\frac{3}{5}}{4/5}\right)=x+c...\left(1\right)$

putting (0,0) pt. we get

$C=\frac{1}{5}ta{n}^{-1}\frac{3}{4}...\left(2\right)$

Putting all values , we get

$\Rightarrow y=\frac{1}{3}ta{n}^{-1}\left[\frac{4ta{n}^{-1}(4x+ta{n}^{-1}\frac{3}{4})-3}{5}\right]-\frac{5x}{3}$