Question

(a) Verify that 3x2 – y2 = c is a one-parameter family of solutions of the differential equation y dy/dx = 3x.

(b) By hand, sketch the graph of the implicit solution 3x2 – y2 = 3. Find all explicit solutions y = f(x) of the DE in part (a) defined by this relation. Give the interval I of definition of each explicit solution.

(c) The point (−2, 3) is on the graph of 3x2 – y2 = 3, but which of the explicit solutions in part (b) satisfies y(−2) = 3?

Short answer

⦁ Proved

b) The four following solutions and their intervals of definition,

$y={\phi }_1\left(x\right)=\sqrt{3}\sqrt{(x+1)(x-1)}$defined on $(1,+\infty )$.

$y={\phi }_2\left(x\right)=-\sqrt{3}\sqrt{(x+1)(x-1)}$defined on $(1,+\infty )$.

$y={\phi }_3\left(x\right)=\sqrt{3}\sqrt{(x+1)(x-1)}$defined on $(-\infty ,-1)$.

$y={\phi }_4\left(x\right)=-\sqrt{3}\sqrt{(x+1)(x-1)}$defined on $(-\infty ,-1)$.

c) The solution is $y=\sqrt{3}\sqrt{(x+1)(x-1)}$

Step-by-step solution

Verifying that 3x2 – y2 = c is a one-parameter family of solutions of the differential equation y dy/dx = 3x

We differentiate our solution,

$$\begin{gathered} 3x^2-y^2=c \\ 6x-2y\frac{dy}{dx}=0 \\ y\frac{dy}{dx}=3x \end{gathered}$$

And we have gotten our differential equation meaning $3x^2-y^2=c$ is the same as saying $y\frac{dy}{dx}=3x$ so it is indeed a family of solutions.

Hence, it is proved that 3x2 – y2 = c is a one-parameter family of solutions of the differential equation y dy/dx = 3x

Finding all explicit solutions

Graph of $3x^2-y^2=3$is shown below,

From $3x^2-y^2=3$we have,

$$\begin{gathered} =\pm \sqrt{3x^2-3} \\ =\pm \sqrt{3}\sqrt{x^2-1} \\ y=\pm \sqrt{3}\sqrt{(x+1)(x-1)} \end{gathered}$$

The solutions are well defined (expression under square root isn’t negative) when either both $x+1$and $x-1$are positive or both $x+1$and $x-1$are negative.

Both $x+1$and $x-1$being positive means $x>-1$and $x>1$so, $x>1$.

Both $x+1$ and $x-1$ being negative means $x<-1$ and $x<1$so, $x<-1$.

That gives us the four following solutions and their intervals of definition,

$y={\phi }_1\left(x\right)=\sqrt{3}\sqrt{(x+1)(x-1)}$defined on$(1,+\infty )$

$y={\phi }_2\left(x\right)=-\sqrt{3}\sqrt{(x+1)(x-1)}$defined on$(1,+\infty )$

$y={\phi }_3\left(x\right)=\sqrt{3}\sqrt{(x+1)(x-1)}$defined on$(-\infty ,-1)$

$y={\phi }_4\left(x\right)=-\sqrt{3}\sqrt{(x+1)(x-1)}$ defined on$(-\infty ,-1)$

Finding which of the explicit solutions in part (b) satisfies y(−2) = 3

We are asked which of ${\phi }_1,{\phi }_2,{\phi }_3,{\phi }_4$can satisfy $y(-2)=3$.

We can immediately discard ${\phi }_1$ and ${\phi }_2$, because they are not even defined for -2.

We can also discard ${\phi }_4$ because it is negative. We left with the only solution that satisfies the condition, ${\phi }_3$

Therefore, the solution is $\mathit{y}\mathbf{=}\sqrt{\mathbf{3}}\sqrt{\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{)}}$