Answer:
[tex]y=5e^{2x}[/tex]
Step-by-step explanation:
Let (x,y) represents a point P on the curve,
So, the slope of the curve at point P = [tex]\frac{dy}{dx}[/tex]
According to the question,
[tex]\frac{dy}{dx}=2y[/tex]
[tex]\frac{1}{y}dy=2dx[/tex]
Integrating both sides,
[tex]\int \frac{dy}{y}=2dx[/tex]
[tex]ln y=2x+ln C[/tex]
[tex]ln y-ln C = 2x[/tex]
[tex]ln(\frac{y}{C})=2x[/tex]
[tex]\frac{y}{C}=e^{2x}[/tex]
[tex]\implies y=Ce^{2x}[/tex]
Since, the curve is passing through the point (0, 5),
[tex]5=Ce^{0}\implies C=5[/tex]
Hence, the required equation of the curve is,
[tex]y=5e^{2x}[/tex]
