Let y=y_1(x) and y=y_2(x) be the solution curves of the differential equation [(d y)/(d x)…

Question

Let $y=y_{1}(x)$ and $y=y_{2}(x)$ be the solution curves of the differential equation $$\frac{d y}{d x}=y+7$$ with initial conditions $y_{1}(0)=0$ and $y_{2}(0)=1$ respectively. Then the curves $y=y_{1}(x)$ and $y=y_{2}(x)$ inte...

Accepted Answer

The given differential equation is

$\frac{dy}{dx}=y+7$

This is a first order linear differential equation and can be solved using an integrating factor.

Rearrange the equation to the standard form of a linear differential equation :

$\frac{dy}{dx}-y=7$

The integrating factor is ${e}^{-\int dx}={e}^{-x}$.

Multiplying each side of the equation by the integrating factor gives :

${e}^{-x}\frac{dy}{dx}-{e}^{-x}y=7{e}^{-x}$

The left-hand side of the equation is the derivative of $({e}^{-x}y)$ with respect to $x$. So we can write the equation as :

$\frac{d}{dx}({e}^{-x}y)=7{e}^{-x}$

Integrate both sides with respect to $x$ :

${e}^{-x}y=-7{e}^{-x}+C$

Multiply both sides by ${e}^{x}$ to isolate $y$ :

$y=-7+C{e}^{x}$

So, the general solution to the differential equation is $y=-7+C{e}^{x}$.

Now, let's apply the initial conditions to find the particular solutions :

For ${y}_{1}(0)=0$, we substitute into the general solution and solve for $C$ :

$0=-7+C$

So, $C=7$, and the solution for ${y}_{1}$ is ${y}_{1}(x)=7{e}^{x}-7$.

For ${y}_{2}(0)=1$, again substitute into the general solution:

$1=-7+C$

So, $C=8$, and the solution for ${y}_{2}$ is ${y}_{2}(x)=8{e}^{x}-7$.

The two curves intersect when ${y}_{1}(x)={y}_{2}(x)$. Setting these equal and solving for $x$ gives :

$7{e}^{x}-7=8{e}^{x}-7$

${e}^{x}=0$

But ${e}^{x}=0$ has no solution, because the exponential function never equals zero.

So, the curves $y={y}_{1}(x)$ and $y={y}_{2}(x)$ do not intersect at any point.

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