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Question

Question: If \(\dfrac{{dy}}{{dx}} = y + 3\) and \(y(0) = 2\) then \(y(\ln (2))\) is equal to: A) \(7\) B) ...

If dydx=y+3\dfrac{{dy}}{{dx}} = y + 3 and y(0)=2y(0) = 2 then y(ln(2))y(\ln (2)) is equal to:
A) 77
B) 55
C) 1313
D) 2 - 2

Explanation

Solution

The given equation is a simple linear differential equation. So we just have to arrange terms of similar variables on one side and then simply integrate to get the function of yy in terms of xx. Then we will use this relation to find desired value.

Complete step-by-step answer:
Given, dydx=y+3\dfrac{{dy}}{{dx}} = y + 3,
After some rearrangements we get,
dyy+3=dx\dfrac{{dy}}{{y + 3}} = dx
Now, Integrate both sides,
dyy+3=dx\int {\dfrac{{dy}}{{y + 3}}} = \int {dx}
Since, dxx=ln(x)+c\int {\dfrac{{dx}}{x} = \ln (x} ) + c and dx=x+c\int {dx = x + c}
So we get,
ln(y+3)+c1=x+c2\ln (y + 3) + {c_1} = x + {c_2}
On further solving we get,
ln(y+3)=x+c2c1\ln (y + 3) = x + {c_2} - {c_1}
This is equivalent to
ln(y+3)=x+c\ln (y + 3) = x + c --------(1)
Now given, y(0)=2y(0) = 2 So putting x=0x = 0 and y=2y = 2 in above equation we get,
ln(2+3)=0+c\ln (2 + 3) = 0 + c
We get cc as,
c=ln(5)c = \ln (5)
Using above value of cc in equation 11 we get final equation as,
ln(y+3)=x+ln(5)\ln (y + 3) = x + \ln (5)
On simplifying, we get,
ln(y+3)ln(5)=x\ln (y + 3) - \ln (5) = x
This simplifies to
ln(y+35)=x\ln (\dfrac{{y + 3}}{5}) = x -------(2)
Because lnalnb=lnab\ln a - \ln b = \ln \dfrac{a}{b}
Now, we want to find value of yy when x=ln(2)x = \ln (2)
So, using equation 22 we get,
ln(y+35)=ln(2)\ln \left( {\dfrac{{y + 3}}{5}} \right) = \ln (2)
On comparing we get,
y+35=2\dfrac{{y + 3}}{5} = 2
On simplification we get,
y+3=10y + 3 = 10
So we get our answer as,
y=7y = 7

Hence, the correct option is (A).

Note: We simplified two constant values in our solutions into a single constant ( c2c1=c{c_2} - {c_1} = c ) because two constants, on subtraction, again gives a constant value which was equivalent to ln(5)\ln (5) . Otherwise, we would not be able to find two constant values separately.