Question

In Problems 11–14 solve the given initial-value problem.

$\mathit{x}{\mathit{y}}^{\mathbf{2}}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}{\mathit{y}}^{\mathbf{3}}\mathbf{-}{\mathit{x}}^{\mathbf{3}}\mathbf{,}\mathit{y}\left(1\right)\mathbf{=}\mathbf{2}$

Short answer

The solution of the given differential equation is $y^3+3x^{3}In\left(x\right)=8x^3$.

Step-by-step solution

Define substitution method for differentiation.

Often, the initial step in solving a differential equation is to use a substitution to change it into another differential equation.

The substitutions that can be employed to solve a homogeneous differential equation are suggested by their properties. A homogeneous equation can be reduced to a separable first-order differential equation by substituting y = ux or x = vy, where u and v are new dependent variables.

Substitution of variables by equations using property.

The given differential equation is a homogeneous function. So, let y = ux. Then,$\frac{dy}{dx}=u+x\frac{du}{dx}$.

Multiply the both sides of the given DE by dx.

$x{y}^{2}dy=\left(y^3-x^3\right)dx$

Substitute them into the differential equation.

$\begin{array}{rcl}x{\left(ux\right)}^2\left(xdu+udx\right)& =& \left({\left(ux\right)}^3-x^3\right)dx\\ u^2{x}^{4}du+u^3{x}^{3}dx& =& u^3{x}^{3}dx-x^{3}dx\\ u^2{x}^{4}du& =& -x^{3}dx\\ u^{2}du& =& -\frac{1}{x}dx\end{array}$

Integrate the function.

Integrate on both sides.

$\begin{array}{rcl}\int u^{2}du& =& -\int \frac{1}{x}dx\\ \frac{1}{3}{u}^3& =& -In\left(x\right)+C\end{array}$

Substitute the equation$u=\frac{y}{x}$into the above equation.

$$\begin{gathered} \frac{y^3}{3x^3}=-In\left(x\right)+C \\ {y}^3=-3x^{3}In\left(x\right)+3x^{3}C \end{gathered}$$

To find the value of constant, substitute y³ (1) = 2³ as y(1) = 2.

$$\begin{gathered} 8=-3{\left(1\right)}^{3}In\left(1\right)+3{\left(1\right)}^{3}C \\ 8=3C \\ C=\frac{8}{3} \end{gathered}$$

Hence, the solution is, $y^3+3x^{3}In\left(x\right)=8x^3$.