Suppose $$𝑦 = 𝑦𝑥$$ be the solution curve to the differential equation $$\frac{dy}{dx}-y=2-e^{-x}$$ such that $$\lim_{x \rightarrow \infty} yx$$ If $$𝑎$$ and $$𝑏$$ are respectively the $$𝑥 -$$ and $$𝑦 -$$ intercept of the tangent to the curve at $$𝑥 = 0$$, then the value of $$𝑎 - 4𝑏$$ is equal to _______.

The differential equation is

$$\frac{dy}{dx}-y=2-e^{-x}$$

The integrating factor is

$$e^{-x}$$

Multiplying throughout by

$$e^{-x}$$

gives

$$e^{-x}\frac{dy}{dx}-e^{-x}y=2e^{-x}-e^{-2x}$$

Hence

$$\frac{d}{dx}(ye^{-x})=2e^{-x}-e^{-2x}$$

Integrating,

$$ye^{-x}=-2e^{-x}+\frac12e^{-2x}+C$$

Multiplying by

$$e^x$$

gives

$$y=-2+\frac12e^{-x}+Ce^x$$

Since

$$\lim_{x\to\infty}y(x)$$

exists and is finite, we must have

$$C=0$$

Therefore

$$y=-2+\frac12e^{-x}$$

At

$$x=0$$

$$y(0)=-2+\frac12=-\frac32$$

Hence

$$b=-\frac32$$

Now differentiate:

$$\frac{dy}{dx}=-\frac12e^{-x}$$

Therefore

$$y'(0)=-\frac12$$

The tangent at

$$x=0$$

is

$$y+\frac32=-\frac12x$$

To find the

$$x$$

-intercept, put

$$y=0$$

$$\frac32=-\frac12x$$

$$x=-3$$

Hence

$$a=-3$$

Therefore

$$a-4b=-3-4\left(-\frac32\right)$$

$$=-3+6$$

$$=3$$

Final Answer :

$$3$$