Question

Find the general solution of the differential equation y'=-x+2^x​

Answer 1

Решение задания приложено

Answer 2

см. прикрепленное изображение ===>>

Answer 3

To find the general solution of the differential equation $y' = -x + 2^x$, we can integrate both sides with respect to $x$.

$\int y' \, dx = \int (-x + 2^x) \, dx$

Integrating the left side with respect to $x$ gives us $y$, and integrating the right side can be done by breaking it into two parts:

$\int (-x + 2^x) \, dx = \int -x \, dx + \int 2^x \, dx$

Integrating each part separately:

$\frac{-x^2}{2} + \int 2^x \, dx$

Now we need to integrate $\int 2^x \, dx$. This can be done by recognizing that the derivative of $2^x$ is $2^x \ln 2$ (by the chain rule). Thus, we have:

$\frac{-x^2}{2} + \int 2^x \, dx = \frac{-x^2}{2} + 2^x \ln 2 + C$

where $C$ is the constant of integration. Therefore, the general solution of the given differential equation is:

$y = \frac{-x^2}{2} + 2^x \ln 2 + C$

where $C$ is an arbitrary constant.