JEE Main 2020 — Differential Equations Question with Solution

Given, $$e^y\left(\frac{dy}{dx}-1\right)=e^x$$

$$\Rightarrow$$ $$e^y\frac{dy}{dx}-e^y=e^x$$ ....(1)

Let $$e^y=t$$ $$\Rightarrow$$ $$e^y\frac{dy}{dx}=\frac{dt}{dx}$$

$$\therefore$$ $$\frac{dt}{dx}-t=e^x$$

So here p = -1 and q = e^x

We know, IF = $$e^{\int pdx}$$

= $$e^{\int -1dx}$$ = $$e^{-x}$$

$$\therefore$$ t.$$e^{-x}$$ = $$\int e^x.e^{-x}dx$$

$$\Rightarrow$$ t.$$e^{-x}$$ = x + c

Putting value of t, we get

$$e^y{e}^{-x}$$ = x + c

$$\Rightarrow$$ $$e^{y-x}=x+c$$ .....(2)

Given y(0) = 0 means y = 0 when x = 0.

Putting in equation (2), we get

e⁰ = 0 + c

$$\Rightarrow$$ c = 1

$$\therefore$$ $$e^{y-x}=x+1$$ ....(3)

Now we have to find y(1), which means when x = 1 find the value of y in the above equation.

Putting x = 1 in equation (3)

$$e^{y-1}=1+1$$

$$\Rightarrow$$ y - 1 = log_e2

$$\Rightarrow$$ y = 1 + log_e2