Question

solve the differential equation dy/dx + y = 2 - x​

Answer 1

Step-by-step explanation:

pls refer to attachments

Answer 2

Answer:

The required solution for the given function is:

$y\times e^x=3e^x-xe^x+c_1$

Or,

$y=\dfrac{3e^x-xe^x+c_1}{e^x}$

Step-by-step explanation:

The given differential equation is:

$\dfrac{dy}{dx}+y=2-x$

The given equation is in the form of the first order differential equation.

i.e:

$y'(x)+P(x)y=Q(x)$

From here we can write the values of:

$P(x)=1\\Q(x)=2-x$

Now solving for the integrating factor we get:

$I.F=e^{\int P(x)\ dx}\\I.F=e^{\int1\ dx}\\I.F=e^x$

Therefore the solution will be:

$y\times I.F=\displaystyle\int Q(x)\times I.F\ dx\\y\times e^x=\displaystyle\int (2-x)e^x\ dx\\y\times e^x=\displaystyle\int 2e^x\ dx-\displaystyle\int xe^x\ dx\,\,\,\,\,eqn(1)$

Now solving the eqn(1) separately we get:

$\displaystyle\int 2e^x\ dx=2e^x+c$

And,

$\displaystyle\int x\ e^x\ dx$

Here we are applying the uv method that has been given below:

$\displaystyle\int (uv)dx=u\displaystyle\int v\ dx-\displaystyle\int [\dfrac{d}{dx}(u)\displaystyle\int v\ dx]dx\\\displaystyle\int xe^x\ dx=x\displaystyle\int e^x\ dx-\displaystyle\int [\dfrac{d}{dx}(x)\displaystyle\int e^x\ dx]dx\\=xe^x-\displaystyle\int [1\cdot e^x]dx\\=xe^x-e^x+c$

Substituting above values in eqn(1) we get the final result:

$y\times e^x=2e^x-xe^x+e^x+c_1$

Or,

$y=\dfrac{3e^x-xe^x+c_1}{e^x}$

Here $c_1$ represents the integration constant.