If d y d x = 2 x + y - 2 x 2 y y ( 0 ) = 1 then y (1) =

Question

If $\frac{d y}{d x} = \frac{2^{x + y} - 2^{x}}{2^{y}},$ $y (0) = 1$ , then $y (1) =$

Smart Achievers · Accepted Answer

(2)

$\frac{d y}{d x} = \frac{2^{x + y} - 2^{x}}{2^{y}} \Rightarrow \frac{d y}{d x} = 2^{x} \frac{(2^{y} - 1)}{2^{y}} \Rightarrow \frac{2^{y}}{2^{y} - 1} d y = 2^{x} d x \Rightarrow \int \frac{2^{y}}{2^{y} - 1} d y = \int 2^{x} d x$

$\Rightarrow \frac{\log (2^{y} - 1)}{\log 2} = \frac{2^{x}}{\log 2} + c, y (0) = 1 \Rightarrow 0 = \frac{1}{\log 2} + c \Rightarrow c = - \frac{1}{\log 2}$

$∴$ $\frac{\log (2^{y} - 1)}{\log 2} = \frac{2^{x}}{\log 2} - \frac{1}{\log 2} \Rightarrow \log (2^{y} - 1) = 2^{x} - 1$

If $x = 1$ than $\log (2^{y} - 1) = 2 - 1 = 1 \Rightarrow 2^{y} - 1 = e \Rightarrow 2^{y} = e + 1 \Rightarrow y = \log_{2} (e + 1)$

$\Rightarrow y (1) = \log_{2} (e + 1)$

Solution

Q.
If $\frac{d y}{d x} = \frac{2^{x + y} - 2^{x}}{2^{y}},$ $y (0) = 1$ , then $y (1) =$

1 $\log_{2} (2 + e)$

2 $\log_{2} (1 + e)$

3 $\log_{2} (2 e^{3})$

4 $\log_{2} (1 + e^{2})$

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Solution

Q. If dydx=2x+y-2x2y, y(0)=1, then y(1)=

1 log2(2+e)

2 log2(1+e)

3 log2(2e3)

4 log2(1+e2)