# Is my solution of the following differential equation wrong?

I had to solve the following differential equation: $(x^2+y)\mathrm{d}x - x \mathrm{d}y=0$. The equation is not exact and so I solved it as a simple linear equation

$$\frac{\mathrm{d}y}{\mathrm{d}x}-\frac{y}{x}=x$$

The solution I got is $\frac{y}{x}=x+c$. However, the solution in the textbook is $-\frac{y}{x}+x=c$. I tried solve it again and again, and yet could not find where I'm wrong. Or is the solution in the book incorrect?

• Congratulations on getting correct answer ! your answer is equivalent to textbook answer. Just notice that : $-y/x+x = c \implies y/x = x-c$ – ganeshie8 Aug 25 '14 at 7:51
• quoting lab bhattacharjee : "Point to be noted, we don't need to match with the answer supplied as long as the approach is right.." – Claude Leibovici Aug 25 '14 at 7:57

Transform your equation with a group $x'=\lambda x$ and $y'=\lambda^\beta y$. Note that $dx'=\lambda dx$ and $dy'=\lambda^\beta dy$.
$$\frac{\lambda^\beta dy}{\lambda dx}-\frac{\lambda^\beta y}{\lambda x}=\lambda x$$For your ODE to be invariant to the group, $\beta -1=1$ or $\beta =2$. Two stabilizers for this group are $$\mu=\frac{y}{x^\beta}=\frac{y}{x^2}$$and $$\nu=\frac{\dot{y}}{x^{\beta -1}}=\frac{\dot{y}}{x}$$where $\dot{y}=\frac{dy}{dx}$. Now your ODE can be written in terms of the group stabilizers: $$\frac{\dot{y}}{x}-\frac{y}{x^2}=1 \rightarrow \nu-\mu=1 \rightarrow \nu=1+\mu$$Noting that $$x\frac{d\mu}{dx}=\nu-\beta \mu=1+\mu-2\mu=1-\mu$$we can separate and solve this equation: $$\frac{d\mu}{1-\mu}=\frac{dx}{x}\rightarrow -ln(1-\mu)=lnx-lnC$$ $$\frac{C}{x}=1-\mu=1-\frac{y}{x^2}\rightarrow Cx=x^2-y$$Thus your answer is equalivent to the one in the textbook: $$y=x^2-Cx$$Of course, since C is a constant it can absorb the negative sign and still work as a solution: $y=x^2+C$.