Question

(a) Show that a one-parameter family of solutions of the equation $\left(4xy+3x^2\right)\mathit{d}\mathit{x}\mathbf{+}\left(2y+2x^2\right)\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$is ${\mathit{x}}^{\mathbf{3}}\mathbf{+}\mathbf{2}{\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathit{c}\mathbf{.}$

(b) Show that the initial conditions$\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{-}\mathbf{}\mathbf{2}$ and $\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{1}$determine the same implicit solution.

(c) Find explicit solutions ${\mathit{y}}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$and${\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ of the differential equation in part (a) such that ${\mathit{y}}_{\mathbf{1}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}$and ${\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}$. Use a graphing utility to graph ${\mathit{y}}_{\mathbf{1}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$and${\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{x}\mathbf{\right)}$.

Short answer

(a) It is shown that the differential equation $\left(3x^2+4xy\right)\mathrm{d}x+\left(2x^2+2y\right)\mathrm{d}y=0$has a solution $x^3+2x^{2}y+y^2=c$.

(b) Both initial conditions determine the same implicit solution $x^3+2x^{2}y+y^2=4$.

(c) The explicit solutions are $y_1\left(x\right)=-x^2-\sqrt{x^4-x^3+4}$and$y_2\left(x\right)=-x^2+\sqrt{x^4-x^3+4}$.

Step-by-step solution

prove the statement in part (a)

Take differential of both sides of the given solution

$\begin{array}{c}\frac{\mathrm{d}}{\mathrm{d}x}\left(x^3+2x^{2}y+y^2\right)=\frac{\mathrm{d}}{\mathrm{d}x}\left(x^3+2x^{2}y+y^2\right)\\ 3x^2+4xy+2x^2\frac{\mathrm{d}y}{\mathrm{d}x}+2y\frac{\mathrm{d}y}{\mathrm{d}x}=0\\ \frac{\mathrm{d}y}{\mathrm{d}x}\left(2x^2+2y\right)=-\left(3x^2+4xy\right)\\ \frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{3x^2+4xy}{2x^2+2y}\end{array}$

Simplify further as

$\begin{array}{c}\left(2x^2+2y\right)\mathrm{d}y=-\left(3x^2+4xy\right)\mathrm{d}x\\ (3x^2+4xy)\mathrm{d}x+\left(2x^2+2y\right)\mathrm{d}y=0\end{array}$

So, the differential equation $\left(3x^2+4xy\right)\mathrm{d}x+\left(2x^2+2y\right)\mathrm{d}y=0$has a solution$x^3+2x^{2}y+y^2=c.$

Obtain the implicit solution in part (b)

Substitute 0 for x, -2 for y into the solution$x^3+2x^{2}y+y^2=c$

$\begin{array}{c}c=(0{)}^3+2(0{)}^2(-2)+(-2{)}^2\\ =4\end{array}$

Now, substitute 1 for x, 1 for y into the solution$x^3+2x^{2}y+y^2=c$

$\begin{array}{rcl}c& =& (1{)}^3+2(1{)}^2\left(1\right)+(1{)}^2\\ c& =& 4\end{array}$

Thus, both initial conditions determine the same implicit solution $x^3+2x^{2}y+y^2=4$.

Obtain the explicit solution in part (c)

Rewrite the solution $x^3+2x^{2}y+y^2=4$as $y^2+2x^{2}y+x^3-4=0$.

The equation $y^2+2x^{2}y+x^3-4=0$is a quadratic equation. The roots of this equation can be calculated as follows

$\begin{array}{c}{y}_{1,2}=\frac{-2x^2\pm \sqrt{{\left(2x^2\right)}^2-4·1·\left(x^3-4\right)}}{2}\\ =\frac{-2x^2\pm \sqrt{4x^4-4x^3+16}}{2}\\ =-x^2\pm \sqrt{x^4-x^3+4}\end{array}$

The solutions $y_1\left(x\right)=-x^2-\sqrt{x^4-x^3+4}$and $y_2\left(x\right)=-x^2+\sqrt{x^4-x^3+4}$are satisfied by $y_1\left(0\right)=-2$and $y_2\left(1\right)=1$respectively as shown in the graph below drawn by using a CAS.