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Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 31

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Answer:y+1=2 e^{\frac{x^{2}}{2}}

Given: The slope of tangent at each point of a curve is equal to the sum of abscissa and product of abscissa and ordinate of the point i.e. \frac{d y}{d x}=x+y

To find: The curves that pass through the point \left ( 0,1 \right )

Hint: Use linear differential equation to find the equation of the curve.

Solution: we have  \frac{d y}{d x}=x+y

        =\frac{d y}{d x}-x y=x \ldots(i)

This is a linear differential equation of the type

=\frac{d y}{d x}-P y=Q

Where P=-x, Q=x

        \begin{aligned} &=I f=e^{\int p d x} \\\\ &=I f=e^{-\int x d x} \\\\ &=I f=e^{-\frac{x^{2}}{2}} \end{aligned}

So solution of equation is given by

        =y e^{-\frac{x^{2}}{2}}=\int \mathrm{x} e^{-\frac{x^{2}}{2}} d x+C \ldots(i i)

\begin{aligned} &\text { Let } I=\int \mathrm{x} e^{-\frac{x^{2}}{2}} d x \\\\ &\text { Let }-\frac{x^{2}}{2}=t \end{aligned}

Differentiating on both sides with respect to x

        \begin{aligned} &=-\frac{2 x}{2} d x=d t \\\\ &=-x d x=d t \\\\ &=x d x=-d t \end{aligned}

        \begin{aligned} &=I=\int e^{-t} d t \\\\ &=I=-e^{-t} \\\\ &=I=-e^{-\frac{x^{2}}{2}} \end{aligned}

Substituting value of I in equation (ii) we get


Dividing by e^{-\frac{x^{2}}{2}}

        =y=-1+C e^{\frac{x^{2}}{2}} \ldots(i i i)

As curve passes through \left ( 0,1 \right )

        \begin{aligned} &=1=-1+C e^{0} \\\\ &=C=2 \end{aligned}

Substituting value of C in equation (iii) we get

        \begin{aligned} &=y=-1+2 e^{\frac{x^{2}}{2}} \\\\ &=y+1=2 e^{\frac{x^{2}}{2}} \end{aligned}

Hence, equation of required curve is found.



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