Question

The solution of the differential equation $\text{log}\left(\frac{dy}{dx}\right)=4x-2y-2$, y = 1 when x = 1 is:

• A. $2e^{2y+2}=e^{4x}+e^2$
• B. $2e^{2y-2}=e^{4x}+e^4$
• C. $2e^{2y+2}=e^{4x}+e^4$
• D. $3e^{2y+2}=e^{3x}+e^4$

Answer 1

The correct option is C

$2e^{2y+2}=e^{4x}+e^4$

$\text{ln}\frac{dy}{dx}=4x-2y-2$

$\frac{dy}{dx}=e^{4x-2y-2}=\frac{e^{4x}}{e^{2y}{e}^2}$

$\therefore \int e^{2y}dy=\int \frac{e^{4x}}{e^2}dx\frac{e^{2y}}{2}=\frac{1}{4}\frac{e^{4x}}{e^2}+c$

For $x=1$, $y=1$

$\frac{e^2}{2}=\frac{1}{4}\frac{e^4}{e^2}+\text{c}\Rightarrow \text{c}=\frac{1}{4}{e}^2$

$\therefore \frac{1}{2}{e}^{2y}=\frac{1}{4}{e}^{4x-2}+\frac{1}{4}{e}^2$

or $2e^{2y}=\frac{e^{4x}}{e^2}+e^2$ or $2e^{2y+2}=e^{4x}+e^4$