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Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.11 question 32 maths

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Answer: 3 y=x^{3}+4

Given: The slope of tangent at each point of a curve is equal to the square of the abscissa of the point i.e. slope of tangent at (x, y)=x^{2}, \frac{d y}{d x}=x+y

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

Hint: First separate the given equation of slope then integrate to find the required curve.

Solution: we have  \frac{d y}{d x}=x^{2}

        =d y=x^{2} d x

Integrate on both sides

        \begin{aligned} &=>\int \mathrm{dy}=\int x^{2} d x \\\\ &=y=\frac{x^{3}}{3}+C \ldots(i) \end{aligned}

As it passes through\left ( -1,1 \right )

        \begin{aligned} &=1=\frac{(-1)^{3}}{3}+C \\\\ &=1=-\frac{1}{3}+C \end{aligned}

         \begin{aligned} &=C=1+\frac{1}{3} \\\\ &=C=\frac{4}{3} \end{aligned}

Substituting value of C in equation (i) we get

        \begin{aligned} &=y=\frac{x^{3}}{3}+\frac{4}{3} \\\\ &=3 y=x^{2}+4 \end{aligned}

Hence, the required curve is found.


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