0
$\begingroup$

I was given a question as follows:

Given that $y = 0$ when $x = 1$, solve the differential equation

$xy(dy/dx)= y^2 + 4,$

obtaining an expression for $y^2$ in terms of $x.$

This is what I did:

enter image description here

However from this it is difficult to eliminate the $\ln y$ and I got stuck...so I went to have a look at the answer.

This is what the first part of the answer said:

Separate variables correctly

Integrate and obtain term $\ln x$

Integrate and obtain term $(1/2)\ln (y^2+4)$

(Each line counts for one mark.)

Now I am really confused... how did they integrate and obtain$ (1/2)\ln(y^2 + 4)$ that suggests $(1/y^2 + 4)$ in the given function and no trace of the $y$ that was after the $x.$

I am probably stupidly overlooking something... but a pointer would be appreciated.

2 Answers 2

2

You made a mistake on the third line where you "turned everything upside down." If you do the fractions properly your RHS should be $$ \frac{y}{y^2 + 4}. $$ Try again with this.

1

That is very easy $$xy(dy/dx)=y^2+4$$ So we have $ydy/(y^2+4)=dx/x $ Integration gives us $$\frac{1}{2} ln(y^2+4)=ln x+c$$ and now we get easily $(y^2+4)^{1/2}=x$ and finally $y^2+4=x^2+c$

  • 0
    You get the integration constant only after integration. $\ln|a|=\ln|b|+c$ has to translate under exponentiation to $a=C·b$ with $C=\pm e^c$. – 2017-01-13
  • 0
    In computing an integral constant values are not important you can get a constant in many cases. – 2017-01-13
  • 1
    Your statement makes no sense. In your first transformation the suddenly appearing $+c$ is wrong. After integration it is correct. – 2017-01-13
  • 0
    Yes you are right I correct it. – 2017-01-13
  • 0
    So you still have to apply the correct exponentiation law to get the correct result. – 2017-01-13