Choose the correct answer from the given four option. Solution of dy dx -y=1, y(0)=1is given by: 1. xy=-e^x 2. xy=-e^ -x 3. xy=-1 4. y=2e^x-1

4. y=2e^x-1 Solution: Given is, dy dx -y=1 dy dx =y+1 dy 1+y =dx On integrating both sides, we get (1+y)=x+C ......(i) When x = 0 and y = 1, then 2=0+C = 2 The required solution is (1+y)=x+ 2 ( 1+y 2 )=x 1+y 2 =e^x 1+y=2e^x =2e^x-1

Choose the correct answer from the given four option. Solution of…

Download App

Question

Choose the correct answer from the given four option.
Solution of $\frac{\text{d}\text{y}}{\text{d}\text{x}}-\text{y}=1,\text{ y}(0)=1$is given by:

$\text{xy}=-\text{e}^\text{x}$
$\text{xy}=-\text{e}^{-\text{x}}$
$\text{xy}=-1$
$\text{y}=2\text{e}^\text{x}-1$

✓

Answer

$\text{y}=2\text{e}^\text{x}-1$

Solution:

Given is, $\frac{\text{d}\text{y}}{\text{d}\text{x}}-\text{y}=1$

$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}=\text{y}+1$

$\Rightarrow\frac{\text{d}\text{y}}{1+\text{y}}=\text{dx}$

On integrating both sides, we get

$\log(1+\text{y})=\text{x}+\text{C}\ ......(\text{i})$

When x = 0 and y = 1, then

$\log2=0+\text{C}$

$\Rightarrow\text{C}=\log2$

The required solution is

$\log(1+\text{y})=\text{x}+\log2$

$\Rightarrow\log\Big(\frac{1+\text{y}}{2}\Big)=\text{x}$

$\Rightarrow\frac{1+\text{y}}{2}=\text{e}^\text{x}$

$\Rightarrow1+\text{y}=2\text{e}^\text{x}$

$\Rightarrow\text{y}=2\text{e}^\text{x}-1$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from DIFFERENTIAL EQUATIONS→DIFFERENTIAL EQUATIONS — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Answer

Need a full question paper?

Explore more

Similar questions