Question

solve each of the following initial value problems

1. (2x-5y)dx+(4x-y)dy=0 y(1)=4
2. (3x²+9xy+5y²)dx-(6x²+4xy)dy=0 y(1)=4

Answer 1

1.

$$(2x-5y)dx+(4x-y)dy=0$$

$$2-5\dfrac{y}{x}+(4-\dfrac{y}{x})\dfrac{dy}{dx}=0$$

Let $y=xv.$ Then $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$

$$2-5v+(4-v)(v+x\dfrac{dv}{dx})=0$$

$$2-5v+4v-v^2+x(4-v)\dfrac{dv}{dx}=0$$

$$\int\dfrac{v-4}{v^2+v-2}dv=-\int\dfrac{dx}{x}$$

$$\dfrac{v-4}{v^2+v-2}=\dfrac{A}{v-1}+\dfrac{B}{v+2}=\dfrac{A(v+2)+B(v-1)}{(v-1)(v+2)}$$

$$A(v+2)+B(v-1)=v-4$$

$$v=1: 3A=-3=>A=-1$$

$$v=-2: -3B=-6=>B=2$$

$$-\int\dfrac{dv}{v-1}+2\int\dfrac{dv}{v+2}=-\int\dfrac{dx}{x}$$

$$-\ln|v-1|+2\ln|v+2|=-\ln|x|+\ln C$$

$$\dfrac{x(v+2)^2}{(v-1)}=C$$

$$\dfrac{(y+2x)^2}{(y-x)}=C$$

$y(1)=4$

$$C=\dfrac{(4+2(1))^2}{(4-1)}=12$$

$$\dfrac{(y+2x)^2}{(y-x)}=12$$

2.

$$(3x^2+9xy+5y^2)dx-(6x^2+4xy)dy=0$$

$$3+9\dfrac{y}{x}+5(\dfrac{y}{x})^2-(6+4\dfrac{y}{x})\dfrac{dy}{dx}=0$$

Let $y=xv.$ Then $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$

$$3+9v+5v^2-(6+4v)(v+x\dfrac{dv}{dx})=0$$

$$3+9v+5v^2-6v-4v^2-x(6+4v)\dfrac{dv}{dx}=0$$

$$\int\dfrac{6+4v}{v^2+3v+3}dv=\int\dfrac{dx}{x}$$

$d(v^2+3v+3)=(2v+3)dv$

$$2\ln(v^2+3v+3)=\ln|x|+\ln C$$

$$\dfrac{(v^2+3v+3)^2}{x}=C$$

$$\dfrac{(y^2+3xy+3x^2)^2}{x^5}=C$$

$y(1)=4$

$$\dfrac{(4^2+3(1)(4)+3(1)^2)^2}{1^5}=C$$

$$C=961$$

$$\dfrac{(y^2+3xy+3x^2)^2}{x^5}=961$$