# General solution formula of a differential equation

1. Nov 8, 2009

### coki2000

Hello,

The general solution of a differential equation for $$y'+P(x)y=G(x)$$ is

$$y(x)=e^{-\int P(x)dx}[C+\int e^{\int P(x)dx}G(x)dx]$$

for $$y'+xy=x$$

$$y(x)=e^{-\int xdx}[C+\int e^{\int xdx}xdx]$$ i have

$$y=Ce^{-\frac{x^2}{2}}+1$$

By the other solution

$$\frac{dy}{dx}+xy=x \rightarrow \frac{dy}{dx}=x(1-y)\Rightarrow \frac{dy}{1-y}=xdx$$

integrating the both sides

$$\int \frac{dy}{1-y}=\int xdx\Rightarrow \ln \left|1-y \right|=\frac{x^2}{2}+C \left|1-y \right|=Ce^{\frac{x^2}{2}}\Rightarrow y=1-Ce^{\frac{x^2}{2}}$$ or $$y=1+Ce^{\frac{x^2}{2}}$$

Why the two solutions are different and where is my mistake?Please help me.Thanks.

2. Nov 8, 2009

### l'Hôpital

when you integrate 1/(1-y), you forgot the negative. It should be - ln (1- y).

3. Nov 8, 2009

### coki2000

ohh thanks l'Hopital

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