Question

The solution of $\frac{dy}{dx}=(3x+y+4{)}^2$ is

• A. $3x+y-4=\sqrt{3}\tan (\sqrt{3x}+c)$
• B. $3x+y+4=\sqrt{3}\tan (x+c)$
• C. $3x+y+4=\sqrt{3}\tan (\sqrt{3}x+c)$
• D. $3x+y+4=\sqrt{3}\tan (2x+c)$

Answer 1

Answer: (A)

$3x+y-4=\sqrt{3}\tan (\sqrt{3x}+c)$

$\frac{dy}{dx}=(3x+y+4{)}^2$

putting $3x+y+4=t$

$\frac{dy}{dx}=\frac{dt}{dx}-3$

So,

$\frac{dt}{dx}-3=t^2$

$\int \frac{dt}{t^2+3}=\int dx$

$\frac{1}{\sqrt{3}}ta{n}^{-1}\frac{t}{\sqrt{3}}=x+C$

Put $t=3x+y+4$

$3x+y+4=\sqrt{3}tan(\sqrt{3}x+C)$