Question

(a) Verify that the one-parameter family${\mathit{y}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathit{x}\mathbf{+}\mathit{c}$is an implicit solution of the differential equation$\mathbf{(}\mathbf{2}\mathit{y}\mathbf{-}\mathbf{2}\mathbf{)}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{2}\mathit{x}\mathbf{-}\mathbf{1}$.

(b) Find a member of the one-parameter family in part (a) that satisfies the initial conditionrole="math" localid="1663826607444" $\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$.

(c) Use your result in part (b) to and an explicit functionrole="math" localid="1663826650077" $\mathit{y}\mathbf{=}\mathit{\varphi }\mathbf{\left(}\mathit{x}\mathbf{\right)}$that satisfies$\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$. Give the domain of the function$\mathit{\varphi }$. Is$\mathit{y}\mathbf{=}\mathit{\varphi }\mathbf{\left(}\mathit{x}\mathbf{\right)}$a solution of the initial-value problem? If so, give its interval Iof definition; if not, explain.

Short answer

(a) $(2y-2)y\text{'}=2x-1$is an implicit solution of the differential equation.

(b)The member of the one-parameter family that satisfies the initial condition$y\left(0\right)=1$is$y^2-2y=x^2-x-1$.

(c) The domain is $x\in ⟨-\infty ,0]\cup [1,\infty ⟩$, and the functions $y=1\pm \sqrt{x^2-x}$are not differentiable at ${\mathrm{x}}_0=0$.

Step-by-step solution

Define the derivation of a function and an explicit function

An explicit function is defined as one that is expressed in terms of the independent variable.

The sensitivity of the function is derived by the derivative of a function of a real variable for modifying the argument.

Step 2(a): Verification of the one-parameter function

Let the given one-parameter function be,$y^2-2y=x^2-x+c$.

Differentiate the given function with respect to.

$\begin{array}{c}\left(y^2-2y\right)\text{'}=\left(x^2-x+c\right)\text{'}\\ 2yy\text{'}-2y\text{'}=2x-1\\ (2y-2)y\text{'}=2x-1\end{array}$

Hence, $(2y-2)y\text{'}=2x-1$is an implicit solution to the differential equation.

Step 3(b): Find the member of the one-parameter family

Substitute the initial value in the given equation and solve for c.

$\begin{array}{c}y(0{)}^2-2y\left(0\right)=0^2-0+c\\ 1-2=c\\ -1=c\end{array}$

Hence, the member of the one-parameter family that satisfies the initial condition $y\left(0\right)=1$is .$y^2-2y=x^2-x-1$.

Step 4(c): Find the domain and the explicit function that satisfies the DE

Solve the given equation to get an explicit function satisfying DE:

$\begin{array}{c}{y}^2-2y+1=x^2-x\\ (y-1{)}^2=x^2-x\\ y=1\pm \sqrt{x^2-x}\end{array}$

Since the function above is defined for all values, except the ones for which the square root is negative, then the following must be satisfied:

$\begin{array}{l}{x}^2-x\ge 0\Rightarrow x(x-1)\ge 0\\ \Rightarrow x\in ⟨-\infty ,0]\cup [1,\infty ⟩\end{array}$

As the functions $y=1\pm \sqrt{x^2-x}$are not differentiable at ${\mathrm{x}}_0=0$, then conclude that neither of the functions $y=1\pm \sqrt{x^2-x}$is not a solution of the initial value problem.

Hence, the domain is $x\in ⟨-\infty ,0]\cup [1,\infty ⟩$, and the functions $y=1\pm \sqrt{x^2-x}$are not differentiable at ${\mathrm{x}}_0=0$.