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

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

Answers (1)

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

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

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.

 

 

Posted by

infoexpert26

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