How to solve :- consider the differential equation dy/dx=1/2x+y-1

Welcome to my article consider the differential equation dy/dx=1/2x+y-1. This question is taken from the simplification lesson.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
For complete information on how to solve this question consider the differential equation dy/dx=1/2x+y-1, read and understand it carefully till the end.

Let us know how to solve this question consider the differential equation dy/dx=1/2x+y-1.

First write the question on the page of the notebook.

consider the differential equation dy/dx=1/2x+y-1 .

\displaystyle \frac{{dy}}{{dx}}=\frac{1}{2}x+y-1

Subtract y from both sides of the equation.

\displaystyle \frac{{dy}}{{dx}}-y=\frac{1}{2}x-1

The integratig factor is definet by the formula of P(x) d(x), where P(x)=-1

\displaystyle {{e}^{{-x}}}

multiply each form by the integrating fector, \displaystyle {{e}^{{-x}}}

\displaystyle {{e}^{{-x}}}\frac{{dy}}{{dx}}+{{e}^{{-x}}}(-y)={{e}^{{-x}}}(\frac{1}{2}x)+{{e}^{{-x}}}\cdot -1

simplify each term ,

rewrite using the commutative propety of multiplication .

\displaystyle {{e}^{{-x}}}\frac{{dy}}{{dx}}+{{e}^{{-x}}}y=\frac{{{{e}^{{-x}}}}}{2}x+{{e}^{{-x}}}\cdot -1

Combine \displaystyle e\frac{{{{e}^{{-x}}}}}{2}and\mathop{x}_{1}

\displaystyle {{e}^{{-x}}}\frac{{dy}}{{dx}}+{{e}^{{-x}}}y=\frac{{{{e}^{{-x}}}x}}{2}+{{e}^{{-x}}}\cdot -1

Move -1 to the left of \displaystyle {{e}^{{-x}}}

\displaystyle {{e}^{{-x}}}\frac{{dy}}{{dx}}+{{e}^{{-x}}}y=\frac{{{{e}^{{-x}}}x}}{2}-1\cdot {{e}^{{-x}}}

\displaystyle {{e}^{{-x}}}\frac{{dy}}{{dx}}-{{e}^{{-x}}}y=\frac{{{{e}^{{-x}}}x}}{2}-1\cdot {{e}^{{-x}}}

Rewrite -1 \displaystyle -1{{e}^{{-x}}}as-e{}^{{-x}}

Reorder factors in \displaystyle {{e}^{{-x}}}\frac{{dy}}{{dx}}-{{e}^{{-x}}}y=\frac{{{{e}^{{-x}}}x}}{2}-{{e}^{{-x}}}

\displaystyle {{e}^{{-x}}}\frac{{dy}}{{dx}}-y{{e}^{{-x}}}=\frac{{x{{e}^{{-x}}}}}{2}-{{e}^{{-x}}}

Rewrite the left as a result of differenatioting aproduct .

\displaystyle \frac{d}{{dx}}-\left[ {{{e}^{{-x}}}y} \right]=\frac{{x{{e}^{{-x}}}}}{2}-{{e}^{{-x}}}

Set up an integraton on each side .

\displaystyle \frac{d}{{dx}}-\left[ {{{e}^{{-x}}}y} \right]dx=\int{{\frac{{x{{e}^{{-x}}}}}{2}-{{e}^{{-x}}}}}dx

Integrate the left side

\displaystyle {{e}^{{-x}}}y=\int{{\frac{{x{{e}^{{-x}}}}}{2}-{{e}^{{-x}}}}}dx

Integrate the right side :-

split the single integrat in to multiple interals

\displaystyle {{e}^{{-x}}}y=\frac{1}{2}\int{{x{{e}^{{-x}}}dx+\int{{{{e}^{{-x}}}}}}}dx

since 1/2 is constant with respect to n ,Move1/2 out of out of the integral

\displaystyle {{e}^{{-x}}}y=\frac{1}{2}\int{{x{{e}^{{-x}}}dx+\int{{-{{e}^{{-x}}}}}}}dx

integrate by parts using the formula , \displaystyle \int{{udv=uv-\int{{vdu}}}},

where u = x and dv = \displaystyle {{e}^{{-x}}}

\displaystyle {{e}^{{-x}}}y=\frac{1}{2}\left( {x\left( {-{{e}^{{-x}}}} \right)-\int{{-{{e}^{{-x}}}dx}}} \right)+\int{{-{{e}^{{-x}}}}}dx

Since -1 is constant with respevt to x, move -1 out of the integral.

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\displaystyle {{e}^{{-x}}}y=\frac{1}{2}\left( {x\left( {-{{e}^{{-x}}}} \right)--\int{{{{e}^{{-x}}}dx}}} \right)+\int{{-{{e}^{{-x}}}}}dx

Simplify

multiply -1 by -1

\displaystyle {{e}^{{-x}}}y=\frac{1}{2}\left( {x\left( {-{{e}^{{-x}}}} \right)+\int{{{{e}^{{-x}}}dx}}} \right)+\int{{-{{e}^{{-x}}}}}dx

multiply \displaystyle \int{{{{e}^{{-x}}}dx}} by 1

\displaystyle {{e}^{{-x}}}y=\frac{1}{2}\left( {x\left( {-{{e}^{{-x}}}} \right)+\int{{{{e}^{{-x}}}dx}}} \right)+\int{{-{{e}^{{-x}}}}}dx

let,

\displaystyle \mathop{u}_{1}=-x

then,

\displaystyle d{{u}_{1}}=-dx

So,

\displaystyle -d{{u}_{1}}=dx

Rewrite using

\displaystyle {{u}_{1}} and \displaystyle d{{u}_{1}}

\displaystyle le{t}'{{u}_{1}}=-x, and \displaystyle d{{u}_{1}}=-dx

\displaystyle So'-d{{u}_{1}}=dx

Rewrite using :-

\displaystyle \mathop{u}_{1}=-x, Find \displaystyle \frac{{\mathop{{du}}_{1}}}{{dx}}

Differentiate -x

\displaystyle \frac{d}{{dx}}\left[ {-x} \right]

Since -1 is constantwith respect to x , the derivative of -x with respect to x is .

\displaystyle -\frac{d}{{dx}}\left[ x \right]

Differentiate using the power rule which states that,

\displaystyle \frac{d}{{dx}}\left[ {{{x}^{n}}} \right]\cdot is\cdot n{{x}^{{n-1}}}

where n=1

multiply -1 by 1

Rewrite the problem using \displaystyle {{u}_{1}} and \displaystyle d{{u}_{1}}

\displaystyle {{e}^{{-x}}}y=\frac{1}{2}(x\left( {-{{e}^{{-x}}}} \right)+ \displaystyle \int{{-{{e}^{{{{u}_{1}}}}}}} \displaystyle {d{{u}_{1}}})+ \displaystyle \int{{-{{e}^{{-x}}}}}dx

Since -1 is constant with respect to \displaystyle \mathop{u}_{1} ,move -1 out of the integral.

\displaystyle {{e}^{{-x}}}y=\frac{1}{2}(x\left( {-{{e}^{{-x}}}} \right)-\int{{{{e}^{{{{u}_{1}}}}}}} \displaystyle d{{u}_{1}}+\int{{-{{e}^{{-x}}}}}dx

The integral of \displaystyle {{e}^{{{{u}_{1}}}}} with respect to \displaystyle {{u}_{1}} is \displaystyle {{e}^{{{{u}_{1}}}}}

\displaystyle {{e}^{{-x}}}y=\frac{1}{2}(x\left( {-{{e}^{{-x}}}} \right)-({{e}^{{{{u}_{1}}}}}+c)+\int{{-{{e}^{{-x}}}}}dx

Since -1 is constant with respect to x, move -1 out of the integral,

\displaystyle {{e}^{{-x}}}y=\frac{1}{2}(x\left( {-{{e}^{{-x}}}} \right)-({{e}^{{{{u}_{1}}}}}+c)+\int{{{{e}^{{-x}}}}}dx

let’ \displaystyle \mathop{u}_{2}=-x

then , \displaystyle \mathop{{du}}_{2}=-dx

so \displaystyle \mathop{{-du}}_{2}=dx

Rewrite using \displaystyle \mathop{u}_{2} and \displaystyle \mathop{{du}}_{2}

let, \displaystyle \mathop{u}_{2} = – x

find, \displaystyle \frac{{\mathop{{du}}_{2}}}{{dx}}

differentiate – x

\displaystyle \frac{d}{{dx}}[-x]

Since -1 is constant with respect to x, the derivative of -x with respect to x is \displaystyle -\frac{d}{{dx}}[x]

\displaystyle -\frac{d}{{dx}}[x]

differentiate using the pouer rule which states that \displaystyle \frac{d}{{dx}}[x] is \displaystyle n{{x}^{{n-1}}}

where n = 1

multiply -i by 1

Rewrite the problem using \displaystyle \mathop{u}_{2} and \displaystyle \mathop{{du}}_{2}

\displaystyle {{e}^{{-x}}}y=\frac{1}{2}(x\left( {-{{e}^{{-x}}}} \right)-({{e}^{{u1}}} \displaystyle +c)+\int{{-{{e}^{{u2}}}}} \displaystyle \underset{2}{\mathop{{du}}}\,

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Since -1 is constant with respect to \displaystyle \mathop{u}_{2}, move -1 out of the integral.

\displaystyle {{e}^{{-x}}}y=\frac{1}{2}(x\left( {-{{e}^{{-x}}}} \right)-({{e}^{{u1}}} \displaystyle +c))--\int{{{{e}^{{u2}}}}} \displaystyle \underset{2}{\mathop{{du}}}\,

Simplity

multiply -1 by 1

\displaystyle {{e}^{{-x}}}y=\frac{1}{2}(x\left( {-{{e}^{{-x}}}} \right)-({{e}^{{u1}}} \displaystyle +c))+ \displaystyle \int{{{{e}^{{u2}}}}} \displaystyle \underset{2}{\mathop{{du}}}\,

multiply \displaystyle \int{{{{e}^{{{{u}_{2}}}}}}} \displaystyle d{{u}_{2}} by 1

\displaystyle {{e}^{{-x}}}y=\frac{1}{2}(x\left( {-{{e}^{{-x}}}} \right)-({{e}^{{\mathop{u}_{1}}}} \displaystyle +c))+\int{{{{e}^{{\mathop{u}_{2}}}}}} \displaystyle \mathop{{du}}_{2}

The integral of \displaystyle \mathop{{du}}_{2} with respect to \displaystyle \mathop{u}_{2} is \displaystyle {{e}^{{\mathop{u}_{2}}}}

\displaystyle {{e}^{{-x}}}y=\frac{1}{2}(x\left( {-{{e}^{{-x}}}} \right)-({{e}^{{u1}}} \displaystyle +c))+\int{{{{e}^{{u2}}}}} \displaystyle +c

Simplity

\displaystyle {{e}^{{-x}}}y=\frac{1}{2}(-x{{e}^{{-x}}}) \displaystyle (-{{e}^{{u1}}})+ \displaystyle {{e}^{{u2}}}+c

Substitute back in for each integration substitute variable replace all occurrences of \displaystyle \mathop{u}_{1} with -x

\displaystyle {{e}^{{-x}}}y=\frac{1}{2}\left( {-x{{e}^{{-x}}}-{{e}^{{-x}}}} \right)+{{e}^{{\mathop{u}_{2}}}}+c

Replace all occurrences of \displaystyle \mathop{u}_{2} -x

\displaystyle {{e}^{{-x}}}y=\frac{1}{2}\left( {-x{{e}^{{-x}}}-{{e}^{{-x}}}} \right)+{{e}^{{-x}}}+c

Simplify

apply the distributive property.

\displaystyle {{e}^{{-x}}}y=\frac{1}{2}\left( {-x{{e}^{{-x}}}} \right)+\frac{1}{2}(-{{e}^{{-x}}})+{{e}^{{-x}}}c

multiply \displaystyle \frac{1}{2}\left( {-x{{e}^{{-x}}}} \right)

Combine 1/2 and x

\displaystyle {{e}^{{-x}}}y=\frac{x}{2}\left( {{{e}^{{-x}}}} \right)+\frac{1}{2}(-{{e}^{{-x}}})+{{e}^{{-x}}}c

Combine \displaystyle {{{e}^{{-x}}}} and x/2

\displaystyle {{e}^{{-x}}}y=-\frac{{{{e}^{{-x}}}x}}{2}+\frac{1}{2}(-{{e}^{{-x}}})+{{e}^{{-x}}}+c

Combine 1/2 and \displaystyle {{{e}^{{-x}}}}

\displaystyle {{e}^{{-x}}}y=\frac{{{{e}^{{-x}}}x}}{2}-\frac{{{{e}^{{-x}}}}}{2}+{{e}^{{-x}}}+c

Reorder factors in \displaystyle -\frac{{{{e}^{{-x}}}x}}{2}

\displaystyle {{e}^{{-x}}}y=\frac{{x{{e}^{{-x}}}}}{2}-\frac{{{{e}^{{-x}}}}}{2}+{{e}^{{-x}}}+c

Reorder terms

\displaystyle {{e}^{{-x}}}y=-\frac{1}{2}x{{e}^{{-x}}}-\frac{1}{2}{{e}^{{-x}}}+{{e}^{{-x}}}+c

Solve for ‘y’

combine x and 1/2

\displaystyle {{e}^{{-x}}}y=-\frac{x}{2}{{e}^{{-x}}}-\frac{1}{2}{{e}^{{-x}}}+{{e}^{{-x}}}+c

Combine \displaystyle {{{e}^{{-x}}}} and x/2

\displaystyle {{e}^{{-x}}}y=-\frac{{{{e}^{{-x}}}x}}{2}-\frac{1}{2}{{e}^{{-x}}}+{{e}^{{-x}}}+c

Divide each term in \displaystyle {{e}^{{-x}}}y

= \displaystyle {{e}^{{-x}}}y=-\frac{{{{e}^{{-x}}}x}}{2}-\frac{{{{e}^{{-x}}}}}{2}+{{e}^{{-x}}}+c

By \displaystyle {{e}^{{-x}}}

y \displaystyle =\frac{{2c-x{{e}^{{-x}}}+{{e}^{{-x}}}}}{{2{{e}^{{-x}}}}}

In this way, after solving this question line by line, its intended solution is =

y \displaystyle =\frac{{2c-x{{e}^{{-x}}}+{{e}^{{-x}}}}}{{2{{e}^{{-x}}}}} (Answer)

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