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### Answers (1)

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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