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.

\(\displaystyle (y-x)y'=y-x+8\) where \(\displaystyle y=x+4\sqrt{x+2}\)

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

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

and the RHS becomes \(\displaystyle y-x+8-x+4\sqrt{x+2}-x+8=4\sqrt{x+2}+8\)
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:

\(\displaystyle y=x+4\sqrt{x+2}\)

then:

\(\displaystyle y'=1+\frac{2}{\sqrt{x+2}}\)

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

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

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

\(\displaystyle y-x+8=x+4\sqrt{x+2}-x+8=4\sqrt{x+2}+8\)

This then shows that:

\(\displaystyle (y-x)y'=y-x+8\)

 

[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.

\(\displaystyle (y-x)y'=y-x+8\) where \(\displaystyle y=x+4\sqrt{x+2}\)

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

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

and the RHS becomes \(\displaystyle y-x+8-x+4\sqrt{x+2}-x+8=4\sqrt{x+2}+8\)
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.

 

[blue_raver22]

In order to verify that y=phi(x) is an explicit solution of the given differential equation, we need to substitute y=phi(x) into the equation and see if it satisfies the equation for all values of x in a certain interval.

In this case, we can take the interval I = [0, ∞) as the interval of definition for the function phi(x) = x+4√(x+2). This means that for all values of x greater than or equal to 0, the function phi(x) is defined and thus can be used as a solution to the differential equation.

Substituting y=phi(x) into the given equation, we get:

(phi(x)-x)(phi'(x)) = phi(x)-x+8

(x+4√(x+2)-x)(1+2/(√(x+2))) = x+4√(x+2)-x+8

4√(x+2)+8 = 4√(x+2)+8

As we can see, the equation is satisfied for all values of x in the interval I = [0, ∞). Therefore, we can conclude that phi(x) = x+4√(x+2) is an explicit solution of the given differential equation.

In general, to verify a solution to a differential equation, we need to substitute the solution into the equation and check if it satisfies the equation for all values in the interval of definition. If it does, then we can say that the solution is valid and can be used to solve the differential equation.