87280
Let \(y=y\) (x) be the solution of the differential equation \((x+1) y^{\prime}-y=e^{3 x}(x+1)^{2}\), with \(y(0)=\frac{1}{3}\). Then, the point \(x=-\frac{4}{3}\) for the curve \(y=y(x)\) is:
1 not a critical point
2 a point of local minima
3 a point of local maxima
4 a point of inflection
Explanation:
(B) : \((x+1) d y-y d x=e^{3 x}(x+1)^{2}\) \(\frac{(x+1) d y-y d x}{(x+1)^{2}}=e^{3 x} \Rightarrow d\left(\frac{y}{x+1}\right)=e^{3 x}\) \(\frac{\mathrm{y}}{\mathrm{x}+1}=\frac{\mathrm{e}^{3 \mathrm{x}}}{3}+\mathrm{C} \Rightarrow\left(0, \frac{1}{3}\right) \Rightarrow \mathrm{C}=0\) \(y=\frac{(x+1) \mathrm{e}^{3 \mathrm{x}}}{3}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{3}\left((\mathrm{x}+1) 3 \mathrm{e}^{3 \mathrm{x}}+\mathrm{e}^{3 \mathrm{x}}\right)=\frac{3^{3 \mathrm{x}}}{3}(3 \mathrm{x}+4)\) Clearly, \(x=\frac{-4}{3}\) is point of local minima
JEE Main-25.06.2022
Differential Equation
87281
Let \(y=y(x)\) be the solution curve of the differential equation \(\frac{d y}{d x}+\frac{1}{x^{2}-1} y=\left(\frac{x-1}{x+1}\right)^{\frac{1}{2}} x>\) 1 passing through the point \(\left(2, \sqrt{\frac{1}{3}}\right)\). Then \(\sqrt{7} \mathrm{y}(8)\) is equal to
1 \(11+6 \log _{\mathrm{e}} 3\)
2 19
3 \(12-2 \log _{\mathrm{e}} 3\)
4 \(19-6 \log _{\mathrm{e}} 3\)
Explanation:
(D) : Given, Curve of differential equation \(\frac{d y}{d x}+\frac{1}{x^{2}-1} y=\left(\frac{x-1}{x+1}\right)^{1 / 2}\) Which is a linear differential equation. I.F \(=\mathrm{e}^{\int \operatorname{Pdx}} \Rightarrow=\mathrm{e}^{\int \frac{1}{\mathrm{x}^{2}-1} \mathrm{dx}} \Rightarrow=\mathrm{e}^{\frac{1}{2} \log \left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)} \Rightarrow \sqrt{\frac{\mathrm{x}-1}{\mathrm{x}+1}}\) For general solution of differential equation \(y \cdot \sqrt{\frac{x-1}{x+1}}=\int \sqrt{\frac{x-1}{x+1}} \cdot \sqrt{\frac{x-1}{x+1}} \cdot d x\) \(y \cdot \sqrt{\frac{x-1}{x+1}}=\int \frac{x-1}{x+1} \cdot d x+C\) \(y \sqrt{\frac{x-1}{x+1}}=\int\left(\frac{x+1-1-1}{x+1}\right) d x\) \(y \sqrt{\frac{x-1}{x+1}}=\int\left(1-\frac{2}{x+1}\right) d x\) \(=x-2 \log (x+1)+C\) Given that curve passing through \((2, \sqrt{1 / 3})\) At, \(\quad \mathrm{x}=2\) \(\sqrt{\frac{1}{3}} \cdot \sqrt{\frac{2-1}{2+1}}=2-2 \log 3+C\) \(\sqrt{\frac{1}{3}} \cdot \sqrt{\frac{1}{3}}=2-2 \log 3+C\) \(C=2 \log 3-2+\frac{1}{3} \Rightarrow C=2 \log 3-5 / 3\) Now become general solution, \(y \sqrt{\frac{x-1}{x+1}}=x-2 \log (x+1)+2 \log 3-5 / 3\) Now find \(\sqrt{7} \mathrm{y}(8)=\) \(\mathrm{y} \sqrt{\frac{8-1}{8+1}}=8-2 \log 9+2 \log 3-5 / 3\) \(\frac{\sqrt{7}}{3} \mathrm{y}=8-2 \log 3-5 / 3\) \(\sqrt{7} \mathrm{y}(8)=24-6 \log 3-5=19-6 \log 3\)
JEE Main-28.07.2022
Differential Equation
87282
The differential equation of the family of circle passing through the points \((0,2)\) and \((0,-2)\) is
1 \(2 x y \frac{d y}{d x}+\left(x^{2}-y^{2}+4\right)=0\)
2 \(2 x y \frac{d y}{d x}+\left(x^{2}+y^{2}-4\right)=0\)
3 \(2 x y \frac{d y}{d x}+\left(y^{2}-x^{2}+4\right)=0\)
4 \(2 x y \frac{d y}{d x}-\left(x^{2}-y^{2}+4\right)=0\)
Explanation:
(A) : The equation of family of circle Passing, \((0,2)\) and \((0,-2)\) \(\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{1}\right)\left(y-y_{2}\right)+\lambda x=0\) \((x-0)(x-0)+(y-2)(y+2)+\lambda(x)=0\) \(x^{2}+y^{2}-4+\lambda x=0 \tag{i}\) \(2 x+2 y \frac{d y}{d x}+\lambda=0 \Rightarrow \lambda=-\left(2 x+2 y \frac{d y}{d x}\right)\) Now putting value of \(\lambda\) in equation (i) \(x^{2}+y^{2}-4-\left(2 x+2 y \frac{d y}{d x}\right) x=0\) \(x^{2}+y^{2}-4-2 x^{2}-2 x y \frac{d y}{d x}=0\) \(-x^{2}+y^{2}-4-2 x y \frac{d y}{d x}=0\) \(2 x y \frac{d y}{d x} x^{2}-y^{2} 40\)
JEE Main-28.07.2022
Differential Equation
87286
The equation of a curve passing through the point \((0,1)\), given that the slope of the tangent to the curve at any point \((x, y)\) is equal to the sum of the \(x\)-coordinate and the product of \(x\) and \(y\) coordinates at that point is
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Differential Equation
87280
Let \(y=y\) (x) be the solution of the differential equation \((x+1) y^{\prime}-y=e^{3 x}(x+1)^{2}\), with \(y(0)=\frac{1}{3}\). Then, the point \(x=-\frac{4}{3}\) for the curve \(y=y(x)\) is:
1 not a critical point
2 a point of local minima
3 a point of local maxima
4 a point of inflection
Explanation:
(B) : \((x+1) d y-y d x=e^{3 x}(x+1)^{2}\) \(\frac{(x+1) d y-y d x}{(x+1)^{2}}=e^{3 x} \Rightarrow d\left(\frac{y}{x+1}\right)=e^{3 x}\) \(\frac{\mathrm{y}}{\mathrm{x}+1}=\frac{\mathrm{e}^{3 \mathrm{x}}}{3}+\mathrm{C} \Rightarrow\left(0, \frac{1}{3}\right) \Rightarrow \mathrm{C}=0\) \(y=\frac{(x+1) \mathrm{e}^{3 \mathrm{x}}}{3}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{3}\left((\mathrm{x}+1) 3 \mathrm{e}^{3 \mathrm{x}}+\mathrm{e}^{3 \mathrm{x}}\right)=\frac{3^{3 \mathrm{x}}}{3}(3 \mathrm{x}+4)\) Clearly, \(x=\frac{-4}{3}\) is point of local minima
JEE Main-25.06.2022
Differential Equation
87281
Let \(y=y(x)\) be the solution curve of the differential equation \(\frac{d y}{d x}+\frac{1}{x^{2}-1} y=\left(\frac{x-1}{x+1}\right)^{\frac{1}{2}} x>\) 1 passing through the point \(\left(2, \sqrt{\frac{1}{3}}\right)\). Then \(\sqrt{7} \mathrm{y}(8)\) is equal to
1 \(11+6 \log _{\mathrm{e}} 3\)
2 19
3 \(12-2 \log _{\mathrm{e}} 3\)
4 \(19-6 \log _{\mathrm{e}} 3\)
Explanation:
(D) : Given, Curve of differential equation \(\frac{d y}{d x}+\frac{1}{x^{2}-1} y=\left(\frac{x-1}{x+1}\right)^{1 / 2}\) Which is a linear differential equation. I.F \(=\mathrm{e}^{\int \operatorname{Pdx}} \Rightarrow=\mathrm{e}^{\int \frac{1}{\mathrm{x}^{2}-1} \mathrm{dx}} \Rightarrow=\mathrm{e}^{\frac{1}{2} \log \left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)} \Rightarrow \sqrt{\frac{\mathrm{x}-1}{\mathrm{x}+1}}\) For general solution of differential equation \(y \cdot \sqrt{\frac{x-1}{x+1}}=\int \sqrt{\frac{x-1}{x+1}} \cdot \sqrt{\frac{x-1}{x+1}} \cdot d x\) \(y \cdot \sqrt{\frac{x-1}{x+1}}=\int \frac{x-1}{x+1} \cdot d x+C\) \(y \sqrt{\frac{x-1}{x+1}}=\int\left(\frac{x+1-1-1}{x+1}\right) d x\) \(y \sqrt{\frac{x-1}{x+1}}=\int\left(1-\frac{2}{x+1}\right) d x\) \(=x-2 \log (x+1)+C\) Given that curve passing through \((2, \sqrt{1 / 3})\) At, \(\quad \mathrm{x}=2\) \(\sqrt{\frac{1}{3}} \cdot \sqrt{\frac{2-1}{2+1}}=2-2 \log 3+C\) \(\sqrt{\frac{1}{3}} \cdot \sqrt{\frac{1}{3}}=2-2 \log 3+C\) \(C=2 \log 3-2+\frac{1}{3} \Rightarrow C=2 \log 3-5 / 3\) Now become general solution, \(y \sqrt{\frac{x-1}{x+1}}=x-2 \log (x+1)+2 \log 3-5 / 3\) Now find \(\sqrt{7} \mathrm{y}(8)=\) \(\mathrm{y} \sqrt{\frac{8-1}{8+1}}=8-2 \log 9+2 \log 3-5 / 3\) \(\frac{\sqrt{7}}{3} \mathrm{y}=8-2 \log 3-5 / 3\) \(\sqrt{7} \mathrm{y}(8)=24-6 \log 3-5=19-6 \log 3\)
JEE Main-28.07.2022
Differential Equation
87282
The differential equation of the family of circle passing through the points \((0,2)\) and \((0,-2)\) is
1 \(2 x y \frac{d y}{d x}+\left(x^{2}-y^{2}+4\right)=0\)
2 \(2 x y \frac{d y}{d x}+\left(x^{2}+y^{2}-4\right)=0\)
3 \(2 x y \frac{d y}{d x}+\left(y^{2}-x^{2}+4\right)=0\)
4 \(2 x y \frac{d y}{d x}-\left(x^{2}-y^{2}+4\right)=0\)
Explanation:
(A) : The equation of family of circle Passing, \((0,2)\) and \((0,-2)\) \(\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{1}\right)\left(y-y_{2}\right)+\lambda x=0\) \((x-0)(x-0)+(y-2)(y+2)+\lambda(x)=0\) \(x^{2}+y^{2}-4+\lambda x=0 \tag{i}\) \(2 x+2 y \frac{d y}{d x}+\lambda=0 \Rightarrow \lambda=-\left(2 x+2 y \frac{d y}{d x}\right)\) Now putting value of \(\lambda\) in equation (i) \(x^{2}+y^{2}-4-\left(2 x+2 y \frac{d y}{d x}\right) x=0\) \(x^{2}+y^{2}-4-2 x^{2}-2 x y \frac{d y}{d x}=0\) \(-x^{2}+y^{2}-4-2 x y \frac{d y}{d x}=0\) \(2 x y \frac{d y}{d x} x^{2}-y^{2} 40\)
JEE Main-28.07.2022
Differential Equation
87286
The equation of a curve passing through the point \((0,1)\), given that the slope of the tangent to the curve at any point \((x, y)\) is equal to the sum of the \(x\)-coordinate and the product of \(x\) and \(y\) coordinates at that point is
87280
Let \(y=y\) (x) be the solution of the differential equation \((x+1) y^{\prime}-y=e^{3 x}(x+1)^{2}\), with \(y(0)=\frac{1}{3}\). Then, the point \(x=-\frac{4}{3}\) for the curve \(y=y(x)\) is:
1 not a critical point
2 a point of local minima
3 a point of local maxima
4 a point of inflection
Explanation:
(B) : \((x+1) d y-y d x=e^{3 x}(x+1)^{2}\) \(\frac{(x+1) d y-y d x}{(x+1)^{2}}=e^{3 x} \Rightarrow d\left(\frac{y}{x+1}\right)=e^{3 x}\) \(\frac{\mathrm{y}}{\mathrm{x}+1}=\frac{\mathrm{e}^{3 \mathrm{x}}}{3}+\mathrm{C} \Rightarrow\left(0, \frac{1}{3}\right) \Rightarrow \mathrm{C}=0\) \(y=\frac{(x+1) \mathrm{e}^{3 \mathrm{x}}}{3}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{3}\left((\mathrm{x}+1) 3 \mathrm{e}^{3 \mathrm{x}}+\mathrm{e}^{3 \mathrm{x}}\right)=\frac{3^{3 \mathrm{x}}}{3}(3 \mathrm{x}+4)\) Clearly, \(x=\frac{-4}{3}\) is point of local minima
JEE Main-25.06.2022
Differential Equation
87281
Let \(y=y(x)\) be the solution curve of the differential equation \(\frac{d y}{d x}+\frac{1}{x^{2}-1} y=\left(\frac{x-1}{x+1}\right)^{\frac{1}{2}} x>\) 1 passing through the point \(\left(2, \sqrt{\frac{1}{3}}\right)\). Then \(\sqrt{7} \mathrm{y}(8)\) is equal to
1 \(11+6 \log _{\mathrm{e}} 3\)
2 19
3 \(12-2 \log _{\mathrm{e}} 3\)
4 \(19-6 \log _{\mathrm{e}} 3\)
Explanation:
(D) : Given, Curve of differential equation \(\frac{d y}{d x}+\frac{1}{x^{2}-1} y=\left(\frac{x-1}{x+1}\right)^{1 / 2}\) Which is a linear differential equation. I.F \(=\mathrm{e}^{\int \operatorname{Pdx}} \Rightarrow=\mathrm{e}^{\int \frac{1}{\mathrm{x}^{2}-1} \mathrm{dx}} \Rightarrow=\mathrm{e}^{\frac{1}{2} \log \left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)} \Rightarrow \sqrt{\frac{\mathrm{x}-1}{\mathrm{x}+1}}\) For general solution of differential equation \(y \cdot \sqrt{\frac{x-1}{x+1}}=\int \sqrt{\frac{x-1}{x+1}} \cdot \sqrt{\frac{x-1}{x+1}} \cdot d x\) \(y \cdot \sqrt{\frac{x-1}{x+1}}=\int \frac{x-1}{x+1} \cdot d x+C\) \(y \sqrt{\frac{x-1}{x+1}}=\int\left(\frac{x+1-1-1}{x+1}\right) d x\) \(y \sqrt{\frac{x-1}{x+1}}=\int\left(1-\frac{2}{x+1}\right) d x\) \(=x-2 \log (x+1)+C\) Given that curve passing through \((2, \sqrt{1 / 3})\) At, \(\quad \mathrm{x}=2\) \(\sqrt{\frac{1}{3}} \cdot \sqrt{\frac{2-1}{2+1}}=2-2 \log 3+C\) \(\sqrt{\frac{1}{3}} \cdot \sqrt{\frac{1}{3}}=2-2 \log 3+C\) \(C=2 \log 3-2+\frac{1}{3} \Rightarrow C=2 \log 3-5 / 3\) Now become general solution, \(y \sqrt{\frac{x-1}{x+1}}=x-2 \log (x+1)+2 \log 3-5 / 3\) Now find \(\sqrt{7} \mathrm{y}(8)=\) \(\mathrm{y} \sqrt{\frac{8-1}{8+1}}=8-2 \log 9+2 \log 3-5 / 3\) \(\frac{\sqrt{7}}{3} \mathrm{y}=8-2 \log 3-5 / 3\) \(\sqrt{7} \mathrm{y}(8)=24-6 \log 3-5=19-6 \log 3\)
JEE Main-28.07.2022
Differential Equation
87282
The differential equation of the family of circle passing through the points \((0,2)\) and \((0,-2)\) is
1 \(2 x y \frac{d y}{d x}+\left(x^{2}-y^{2}+4\right)=0\)
2 \(2 x y \frac{d y}{d x}+\left(x^{2}+y^{2}-4\right)=0\)
3 \(2 x y \frac{d y}{d x}+\left(y^{2}-x^{2}+4\right)=0\)
4 \(2 x y \frac{d y}{d x}-\left(x^{2}-y^{2}+4\right)=0\)
Explanation:
(A) : The equation of family of circle Passing, \((0,2)\) and \((0,-2)\) \(\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{1}\right)\left(y-y_{2}\right)+\lambda x=0\) \((x-0)(x-0)+(y-2)(y+2)+\lambda(x)=0\) \(x^{2}+y^{2}-4+\lambda x=0 \tag{i}\) \(2 x+2 y \frac{d y}{d x}+\lambda=0 \Rightarrow \lambda=-\left(2 x+2 y \frac{d y}{d x}\right)\) Now putting value of \(\lambda\) in equation (i) \(x^{2}+y^{2}-4-\left(2 x+2 y \frac{d y}{d x}\right) x=0\) \(x^{2}+y^{2}-4-2 x^{2}-2 x y \frac{d y}{d x}=0\) \(-x^{2}+y^{2}-4-2 x y \frac{d y}{d x}=0\) \(2 x y \frac{d y}{d x} x^{2}-y^{2} 40\)
JEE Main-28.07.2022
Differential Equation
87286
The equation of a curve passing through the point \((0,1)\), given that the slope of the tangent to the curve at any point \((x, y)\) is equal to the sum of the \(x\)-coordinate and the product of \(x\) and \(y\) coordinates at that point is
87280
Let \(y=y\) (x) be the solution of the differential equation \((x+1) y^{\prime}-y=e^{3 x}(x+1)^{2}\), with \(y(0)=\frac{1}{3}\). Then, the point \(x=-\frac{4}{3}\) for the curve \(y=y(x)\) is:
1 not a critical point
2 a point of local minima
3 a point of local maxima
4 a point of inflection
Explanation:
(B) : \((x+1) d y-y d x=e^{3 x}(x+1)^{2}\) \(\frac{(x+1) d y-y d x}{(x+1)^{2}}=e^{3 x} \Rightarrow d\left(\frac{y}{x+1}\right)=e^{3 x}\) \(\frac{\mathrm{y}}{\mathrm{x}+1}=\frac{\mathrm{e}^{3 \mathrm{x}}}{3}+\mathrm{C} \Rightarrow\left(0, \frac{1}{3}\right) \Rightarrow \mathrm{C}=0\) \(y=\frac{(x+1) \mathrm{e}^{3 \mathrm{x}}}{3}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{3}\left((\mathrm{x}+1) 3 \mathrm{e}^{3 \mathrm{x}}+\mathrm{e}^{3 \mathrm{x}}\right)=\frac{3^{3 \mathrm{x}}}{3}(3 \mathrm{x}+4)\) Clearly, \(x=\frac{-4}{3}\) is point of local minima
JEE Main-25.06.2022
Differential Equation
87281
Let \(y=y(x)\) be the solution curve of the differential equation \(\frac{d y}{d x}+\frac{1}{x^{2}-1} y=\left(\frac{x-1}{x+1}\right)^{\frac{1}{2}} x>\) 1 passing through the point \(\left(2, \sqrt{\frac{1}{3}}\right)\). Then \(\sqrt{7} \mathrm{y}(8)\) is equal to
1 \(11+6 \log _{\mathrm{e}} 3\)
2 19
3 \(12-2 \log _{\mathrm{e}} 3\)
4 \(19-6 \log _{\mathrm{e}} 3\)
Explanation:
(D) : Given, Curve of differential equation \(\frac{d y}{d x}+\frac{1}{x^{2}-1} y=\left(\frac{x-1}{x+1}\right)^{1 / 2}\) Which is a linear differential equation. I.F \(=\mathrm{e}^{\int \operatorname{Pdx}} \Rightarrow=\mathrm{e}^{\int \frac{1}{\mathrm{x}^{2}-1} \mathrm{dx}} \Rightarrow=\mathrm{e}^{\frac{1}{2} \log \left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)} \Rightarrow \sqrt{\frac{\mathrm{x}-1}{\mathrm{x}+1}}\) For general solution of differential equation \(y \cdot \sqrt{\frac{x-1}{x+1}}=\int \sqrt{\frac{x-1}{x+1}} \cdot \sqrt{\frac{x-1}{x+1}} \cdot d x\) \(y \cdot \sqrt{\frac{x-1}{x+1}}=\int \frac{x-1}{x+1} \cdot d x+C\) \(y \sqrt{\frac{x-1}{x+1}}=\int\left(\frac{x+1-1-1}{x+1}\right) d x\) \(y \sqrt{\frac{x-1}{x+1}}=\int\left(1-\frac{2}{x+1}\right) d x\) \(=x-2 \log (x+1)+C\) Given that curve passing through \((2, \sqrt{1 / 3})\) At, \(\quad \mathrm{x}=2\) \(\sqrt{\frac{1}{3}} \cdot \sqrt{\frac{2-1}{2+1}}=2-2 \log 3+C\) \(\sqrt{\frac{1}{3}} \cdot \sqrt{\frac{1}{3}}=2-2 \log 3+C\) \(C=2 \log 3-2+\frac{1}{3} \Rightarrow C=2 \log 3-5 / 3\) Now become general solution, \(y \sqrt{\frac{x-1}{x+1}}=x-2 \log (x+1)+2 \log 3-5 / 3\) Now find \(\sqrt{7} \mathrm{y}(8)=\) \(\mathrm{y} \sqrt{\frac{8-1}{8+1}}=8-2 \log 9+2 \log 3-5 / 3\) \(\frac{\sqrt{7}}{3} \mathrm{y}=8-2 \log 3-5 / 3\) \(\sqrt{7} \mathrm{y}(8)=24-6 \log 3-5=19-6 \log 3\)
JEE Main-28.07.2022
Differential Equation
87282
The differential equation of the family of circle passing through the points \((0,2)\) and \((0,-2)\) is
1 \(2 x y \frac{d y}{d x}+\left(x^{2}-y^{2}+4\right)=0\)
2 \(2 x y \frac{d y}{d x}+\left(x^{2}+y^{2}-4\right)=0\)
3 \(2 x y \frac{d y}{d x}+\left(y^{2}-x^{2}+4\right)=0\)
4 \(2 x y \frac{d y}{d x}-\left(x^{2}-y^{2}+4\right)=0\)
Explanation:
(A) : The equation of family of circle Passing, \((0,2)\) and \((0,-2)\) \(\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{1}\right)\left(y-y_{2}\right)+\lambda x=0\) \((x-0)(x-0)+(y-2)(y+2)+\lambda(x)=0\) \(x^{2}+y^{2}-4+\lambda x=0 \tag{i}\) \(2 x+2 y \frac{d y}{d x}+\lambda=0 \Rightarrow \lambda=-\left(2 x+2 y \frac{d y}{d x}\right)\) Now putting value of \(\lambda\) in equation (i) \(x^{2}+y^{2}-4-\left(2 x+2 y \frac{d y}{d x}\right) x=0\) \(x^{2}+y^{2}-4-2 x^{2}-2 x y \frac{d y}{d x}=0\) \(-x^{2}+y^{2}-4-2 x y \frac{d y}{d x}=0\) \(2 x y \frac{d y}{d x} x^{2}-y^{2} 40\)
JEE Main-28.07.2022
Differential Equation
87286
The equation of a curve passing through the point \((0,1)\), given that the slope of the tangent to the curve at any point \((x, y)\) is equal to the sum of the \(x\)-coordinate and the product of \(x\) and \(y\) coordinates at that point is
