Verifying solution to first order differential equation

[find_the_fun]

Verify the indicated function y=phi(x) is an explicit solution of the given equation. Consider the phi function as a solution of the differential equation and give at lease one interval I of definition.

[math](y-x)y'=y-x+8[/math] where [math]y=x+4\sqrt{x+2}[/math]

So the derivative is [math]y'=1+\frac{2}{\sqrt{x+2}}[/math]
and the LHS becomes [math](y-x)(1+\frac{2}{\sqrt{x+2}})=
y+\frac{2y}{\sqrt{x+2}}-x-\frac{2x}{\sqrt{x+2}}=[/math]

$$x+4\sqrt{x+2}+2(x+\frac{4 \sqrt{x+2}}{\sqrt{x+2}})-x-\frac{2}{\sqrt{x+2}}=4\sqrt{x+2}+2x+8-\frac{2x}{\sqrt{x+2}}$$

and the RHS becomes [math]y-x+8-x+4\sqrt{x+2}-x+8=4\sqrt{x+2}+8[/math]
which isn't equal but the answer key seems to think it is because it gives an interval of definition.

 

[MarkFL]

I agree that if:

$$y=x+4\sqrt{x+2}$$

then:

$$y'=1+\frac{2}{\sqrt{x+2}}$$

Now, let's look at the left side of the ODE:

$$(y-x)y'=\left(x+4\sqrt{x+2}-x \right)\left(1+\frac{2}{\sqrt{x+2}} \right)=4\sqrt{x+2}+8$$

And next, let's look at the right side:

$$y-x+8=x+4\sqrt{x+2}-x+8=4\sqrt{x+2}+8$$

This then shows that:

$$(y-x)y'=y-x+8$$

 

[Prove It]

Verify the indicated function y=phi(x) is an explicit solution of the given equation. Consider the phi function as a solution of the differential equation and give at lease one interval I of definition.

[math](y-x)y'=y-x+8[/math] where [math]y=x+4\sqrt{x+2}[/math]

So the derivative is [math]y'=1+\frac{2}{\sqrt{x+2}}[/math]
and the LHS becomes [math](y-x)(1+\frac{2}{\sqrt{x+2}})=
y+\frac{2y}{\sqrt{x+2}}-x-\frac{2x}{\sqrt{x+2}}=[/math]

$$x+4\sqrt{x+2}+2(x+\frac{4 \sqrt{x+2}}{\sqrt{x+2}})-x-\frac{2}{\sqrt{x+2}}=4\sqrt{x+2}+2x+8-\frac{2x}{\sqrt{x+2}}$$

and the RHS becomes [math]y-x+8-x+4\sqrt{x+2}-x+8=4\sqrt{x+2}+8[/math]
which isn't equal but the answer key seems to think it is because it gives an interval of definition.

\displaystyle \begin{align*} \left( y - x \right) \, \frac{dy}{dx} &= y - x + 8 \end{align*}

If \displaystyle \begin{align*} y = x + 4\sqrt{x+2} \end{align*} then \displaystyle \begin{align*} \frac{dy}{dx} = 1 + \frac{2}{\sqrt{x + 2}} \end{align*} and so substituting into the DE we have:

\displaystyle \begin{align*} LHS &= \left( y - x \right) \, \frac{dy}{dx} \\ &= \left( x + 4\sqrt{x + 2} - x \right) \left( 1 + \frac{2}{\sqrt{x + 2}} \right) \\ &= x + \frac{2x}{\sqrt{x + 2}} + 4\sqrt{x + 2} + 8 - x - \frac{2x}{\sqrt{x + 2}} \\ &= \left( x + 4\sqrt{x + 2} \right) - x + 8 \\ &= y - x + 8 \\ &= RHS \end{align*}

 

[Prove It]

Another possibility is brute force, actually solving the DE:

\displaystyle \begin{align*} \left( y - x \right) \, \frac{dy}{dx} &= y - x + 8 \\ \frac{dy}{dx} &= \frac{y - x + 8}{y - x} \end{align*}

Make the substitution \displaystyle \begin{align*} u = y - x \implies \frac{du}{dx} = \frac{dy}{dx} - 1 \implies \frac{dy}{dx} = \frac{du}{dx} + 1 \end{align*} and the DE becomes

\displaystyle \begin{align*} \frac{dy}{dx} &= \frac{y - x + 8}{y - x} \\ \frac{du}{dx} + 1 &= \frac{u + 8}{u} \\ \frac{du}{dx} &= \frac{u + 8}{u} - 1 \\ \frac{du}{dx} &= \frac{8}{u} \\ u\,\frac{du}{dx} &= 8 \\ \int{ u\,\frac{du}{dx}\,dx} &= \int{8\,dx} \\ \int{u\,du} &= 8x + C_1 \\ \frac{1}{2}u^2 + C_2 &= 8x + C_1 \\ \frac{1}{2}u^2 &= 8x + C_1 - C_2 \\ u^2 &= 16x + C \textrm{ where } C = 2C_1 - 2C_2 \\ \left( y - x \right) ^2 &= 16x + C \\ y - x &= \sqrt{16x + C} \\ y &= x + \sqrt{16x + C} \end{align*}

Of course, your given function \displaystyle \begin{align*} y = x + 4\sqrt{x + 2} \end{align*} is the case where \displaystyle \begin{align*} C = 32 \end{align*}, thus your given function is a solution to the DE.