Question

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

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

Short answer

The solution of the given differential equation is y² = 2x⁴ - x².

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,$\frac{dy}{dx}=u+x\frac{du}{dx}$.

Substitute them into the differential equation.

$\begin{array}{rcl}\left(x^2+2y^2\right)\frac{dx}{dy}& =& xy\\ xy\frac{dy}{dx}& =& x^2+2y^2\\ x\left[ux\right]\left[u+x\frac{du}{dx}\right]& =& x^2+2{\left[ux\right]}^2\\ u^2{x}^2+u{x}^3\frac{du}{dx}& =& x^2+2u^2{x}^2\\ u{x}^3\frac{du}{dx}& =& x^2+u^2{x}^2\\ u{x}^3\frac{du}{dx}& =& x^2\left(1+u^2\right)\\ ux\frac{du}{dx}& =& \left(1+u^2\right)\end{array}$

Integrate the function.

Let separate the variables be,

$\frac{u}{u^2+1}du=\frac{1}{x}dx$

Integrate on both sides.

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

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

$\begin{array}{rcl}{\left(\frac{y}{x}\right)}^2& =& C{x}^2-1\\ y^2& =& C{x}^4-x^2\end{array}$

To find the value of constant, substitute y (- 1) = 1.

1 = C - 1

C = 2

Hence, the solution is, y² = 2x⁴ - x².