Question

I need help to solve this
dy/dx=xy+x-2y-2 // y(0)=2

thank you

Answer 1

Hi tonijv,

This is a separable equation. Factorize the right-hand side to see,\[y(x-2)+(x-2)=(y+1)(x-2)\]You then have,\[\frac{dy}{y+1}=(x-2)dx \rightarrow \int\limits_{}^{} \frac{dy}{y+1}=\int\limits_{}^{}(x-2)dx\]

Answer 2

\[\ln(y+1)=\frac{x^2}{2}-2x+c \rightarrow y=e^{\frac{x^2}{2}-2x+c}-1\]That is,\[y=Ae^{\frac{x^2}{2}-2x}-1\]Using the boundary condition,\[y(0)=Ae^0-1=2\rightarrow A=3\]

Answer 3

Your equation is then,\[y=3e^{\frac{x^2}{2}-2x}\]

Answer 4

The previous comment was correct. This is a separable equation and wanted to also note a slightly different way of factorizing the RHS:

\[dy/dx = xy + x - 2y - 2
= x(y + 1) - 2(y+1)
= (y+1) (x-2)\]

Now place all of the y variables to one side and all of the x variables to the other and you obtain the integral that lokisan noted above. Upon solving this integral you should get:
\[\ln \left| y+1 \right|=x ^{2}/2 - 2x + C\]
Now plug in the initial value conditions for x = 0 and y = 1 and solve for C to obtain: \[C=\ln(2)\]
Thereby, making your final answer:
\[ \ln \left| y+1 \right|=x ^{2}/2 - 2x + \ln(2)\]

Answer 5

thanks guys is my first time here. I am lost in differentials. greetings from Barcelona

Answer 6

No probs.

Answer 7

I have given you the answer "implicitly" in terms of y but you can also solve this "explicitly" for y, but some professors don't mind which form you use.

Answer 8

\[y=(1/2)(x-4)(x)(y+1)+2\]
The derivative of the right side of the above with respect to x is:
\[(1/2)(x-4)(y+1)+(1/2) (x)(y+1)\]
or
\[x y+x-2 y-2\]