Answer to Question #218932 in Differential Equations for maxine

Question

solve

dy/dx+4xy=8x

(6xy+2y2-5)dx+(3x2+4xy-6)dy=0 y(1)=2

Answer 1

(a) Solve the separable equation

$$\frac{d y(x)}{d x}+4 x y(x)=8x :$$

Solve for $\frac{d y(x)}{d x} :$ :

$$\frac{d y(x)}{d x}=-4(-2 x+x y(x))$$

Simplify:

$$\frac{d y(x)}{d x}=x(-4 y(x)+8)$$

Divide both sides by $-4 y(x)+8:$

$$\frac{\frac{d y(x)}{d x}}{-4 y(x)+8}=x$$

Integrate both sides with respect to $\boldsymbol{x}:$

$$\int \frac{\frac{d y(x)}{d x}}{-4 y(x)+8} d x= \int x d x$$

Evaluating this integrals:

$$-\frac{1}{4} \log (-4 y(x)+8)=\frac{x^{2}}{2}+c_{1}$$

where $c_{1}$ is an arbitrary constant

Solve for $y(x):$

$$y(x)=-\frac{1}{4} e^{-2\left(x^{2}+2 c_{1}\right)}+2$$

By simplifying the arbitrary constants, we have:

$$y(x)=-\frac{1}{4} e^{-2\left(x^{2}+c_{1}\right)}+2$$

(b)

$$(6xy+2y^2-5)dx+(3x^2+4xy-6)dy=0 \qquad y(1)=2$$

The given differential equation is of the form:

$$Mdx + Ndy =0$$

With:

$$M = 6xy+2y^2-5 \text{ and } N= 3x^2+4xy-6$$

We proceed to check the exactness of the two equations. For exactness,

$$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$

Thus:

$$\frac{\partial M}{\partial y}=6x+4y=\frac{\partial N}{\partial x}$$

We then proceed to solve the equation.

$$\Phi_x = \int(6xy+2y^2-5)dx = 3x^2y+2xy^2-5x\\ \Phi_y= \int (3x^2+4xy-6)dy= 3x^2y+2xy^2-6y$$

The general solution of the DE is:

$$3x^2y+2xy^2-5x-6y=c$$

Evaluating the general solution using the IVP $y(1)=2$ :

$$3(1)^2(2)+2(1)(2)^2-5(1)-6(2)=c\\ 6+8-5-12=c\\ \therefore c=-3$$

Thus the solution is:

$$3x^2y+2xy^2-5x-6y=-3$$