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  • Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 30

Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 30

Answers (1)

Answer: x+y=e^{x}-1

Given: The slope of tangent at each point of a curve is equal to the sum of the co-ordinate of the point i.e. slope of tangent at (x, y)=x+y

To find: The curves that pass through the origin.

Hint: As  \frac{d y}{d x}+P y=Q then it is a linear differential equation.

Solution: Slope of tangent at (x, y)=x+y

        \begin{aligned} &=\frac{d y}{d x}=x+y \\\\ &=\frac{d y}{d x}-y=x \end{aligned}

It is a linear differential equation

Comparing it with  \frac{d y}{d x}+P y=Q

        =P=-1, Q=x

        =I f=e^{\int \mathrm{P} d x}

        \begin{aligned} &=I f=e^{-\int 1 d x} \\\\ &=I f=e^{-x} \end{aligned}

It is a linear differential equation

Comparing it with  \frac{d y}{d x}+P y=Q

        \begin{aligned} &=y(I f)=\int \mathrm{Q}(I f) d x+C \\\\ &=y\left(e^{-x}\right)=\int x e^{-x} d x+C \end{aligned}

        =y e^{-x}=x \int e^{-x} d x-\left(\frac{d x}{d x} \int e^{-x} d x\right) d x+C        [using integration by parts]

        \begin{aligned} &=y e^{-x}=-x e^{-x}-\int e^{-x} d x+C \\\\ &=y e^{-x}=-x e^{-x}+e^{-x}+C \end{aligned}

Dividing by e^{-x} on both sides

        =y=-x-1+C e^{x} \ldots(i)

As it passes through origin (x, y)=(0,0)

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

Substituting C in equation (i)

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

Hence, equation of curve is found

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