Math, asked by nerdy27, 10 months ago

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

Answers

Answered by ferozemulani
1

Step-by-step explanation:

pls refer to attachments

Attachments:
Answered by shalusingh583
1

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.

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