The solution of the differential equation (1+x) d y/d x-x y=1-x is

Correct Answer is : y(1+x)=x+cexy(1+x)=x+{ce}^{x}y(1+x)=x+cex, where c is the constant of integration

y(1+x)={ce}^{x}, where c is the constant of integration

y(1+x)=x, where c is the constant of integration

Correct Answer is : y(1+x)=x+cexy(1+x)=x+{ce}^{x}y(1+x)=x+cex, where c is the constant of integration

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MHTCET 26 Apr 2025 Shift 2 Paper

Question : 120 of 150

Marks: +1, -0

(1+x) \frac{d y}{d x}-x y=1-x

Solution:

Concept:

This is a first-order linear differential equation, solved using an integrating factor.

Explanation:

Given:

(1+x)\frac{dy}{dx}-xy=1-x

Rewrite in standard linear form:

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

Integrating factor:

I.F.

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

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

=e^{-x+\ln(1+x)}

=e^{-x}(1+x)

Multiply both sides of the equation by the I.F.:

y(1+x)e^{-x}=\int \frac{1-x}{1+x}(1+x)e^{-x}dx+c

y(1+x)e^{-x}=\int(1-x)e^{-x}dx+c

=\int e^{-x}dx-\int xe^{-x}dx+c

=-e^{-x}-\left(-xe^{-x}-e^{-x}\right)+c

=xe^{-x}+c

Thus:

y(1+x)e^{-x}=xe^{-x}+c

Multiplying by

e^x

y(1+x)=x+ce^x

Answer:

y(1+x)=x+ce^x

, where

c

is the constant of integration.

Therefore, the correct option is A.

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