Differential Equations
Substitution to reduce to separable form
MMTS_Full_Test_17
Grade 12
Question:
The solution of the differential equation $\dfrac{dy}{dx}=(4x+y+1)^2$ is
(A) $4x+y+1=2\tan(2x+y+C)$... wait — see below
(B) $4x+y+1=2\tan(x+2y+C)$
(C) $4x+y+1=2\tan(2y+C)$
(D) $4x+y+1=2\tan(2x+C)$
Step-by-Step Solution
Key Concept: Let $v=4x+y+1$; $dv/dx=4+dy/dx=4+v^2$. Separate: $dv/(4+v^2)=dx$. Integrate: $\frac{1}{2}\tan^{-1}(v/2)=x+C'$, so $v=2\tan(2x+C)$.
$4x+y+1=2\tan(2x+C)$.
Correct Answer: (D) $4x+y+1=2\tan(2x+C)$