Answer to Question #260746 in Differential Equations for christian

Question

1. ( y-2x)dx+xdy=0 ,y(-1)=3
2. dy=(x-4xy)dx ,y(0)= - 1/4

Answer 1

Let us solve the following differential equations.

1. $(y-2x)dx+xdy=0 ,y(-1)=3$

This equation can be rewritten in the form $xy'+y-2x=0$ or $(xy)'=2x.$ It follows that $xy=x^2+C,$ and we conclude that the general solution is of the form: $y=x+\frac{C}x.$ Then $3=y(-1)=-1-C,$ and thus $C=-4.$ The solution of $(y-2x)dx+xdy=0 ,y(-1)=3,$ is $y=x-\frac{4}x.$

2. $dy=(x-4xy)dx ,y(0)= - 1/4$

This equation is equivalent to $\frac{dy}{1-4y}=xdx,$ and hence $\int\frac{dy}{1-4y}=\int xdx.$ Then $-\frac{1}4\ln|1-4y|=\frac{x^2}2+C.$

It follows that $y=-\frac{1}4$ when $x=0,$ and hence $-\frac{1}4\ln 2=C.$

We conclude that the solution of $dy=(x-4xy)dx ,y(0)= - 1/4$

is of the form $-\frac{1}4\ln|1-4y|=\frac{x^2}2-\frac{1}4\ln 2.$