- #1
coki2000
- 91
- 0
Hello,
The general solution of a differential equation for [tex]y'+P(x)y=G(x)[/tex] is
[tex]y(x)=e^{-\int P(x)dx}[C+\int e^{\int P(x)dx}G(x)dx][/tex]
for [tex]y'+xy=x[/tex]
[tex]y(x)=e^{-\int xdx}[C+\int e^{\int xdx}xdx][/tex] i have
[tex]y=Ce^{-\frac{x^2}{2}}+1[/tex]
By the other solution
[tex]\frac{dy}{dx}+xy=x \rightarrow \frac{dy}{dx}=x(1-y)\Rightarrow \frac{dy}{1-y}=xdx[/tex]
integrating the both sides
[tex]\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}}[/tex] or [tex]y=1+Ce^{\frac{x^2}{2}}[/tex]
Why the two solutions are different and where is my mistake?Please help me.Thanks.
The general solution of a differential equation for [tex]y'+P(x)y=G(x)[/tex] is
[tex]y(x)=e^{-\int P(x)dx}[C+\int e^{\int P(x)dx}G(x)dx][/tex]
for [tex]y'+xy=x[/tex]
[tex]y(x)=e^{-\int xdx}[C+\int e^{\int xdx}xdx][/tex] i have
[tex]y=Ce^{-\frac{x^2}{2}}+1[/tex]
By the other solution
[tex]\frac{dy}{dx}+xy=x \rightarrow \frac{dy}{dx}=x(1-y)\Rightarrow \frac{dy}{1-y}=xdx[/tex]
integrating the both sides
[tex]\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}}[/tex] or [tex]y=1+Ce^{\frac{x^2}{2}}[/tex]
Why the two solutions are different and where is my mistake?Please help me.Thanks.