The equation of the curve passing through (1, | vedclass

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(A) Given point $P=(1,2)$.The slope of the tangent is given by $\frac{dy}{dx} = 	an(	heta) = 	an(	an^{-1}(2x+3y)) = 2x+3y$.This is a linear differ

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$6 x+9 y+2=26 e^{3 x-3}$

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(A) Given point $P=(1,2)$.The slope of the tangent is given by $\frac{dy}{dx} = 	an(	heta) = 	an(	an^{-1}(2x+3y)) = 2x+3y$.This is a linear differential equation: $\frac{dy}{dx} - 3y = 2x$.Integrating factor $IF = e^{\int -3 dx} = e^{-3x}$.Multiplying both sides by $IF$: $y e^{-3x} = \int 2x e^{-3x} dx + c$.Using integration by parts: $\int 2x e^{-3x} dx = 2x \left(\frac{e^{-3x}}{-3}ight) - \int 2 \left(\frac{e^{-3x}}{-3}ight) dx = -\frac{2}{3}x e^{-3x} - \frac{2}{9} e^{-3x} + c$.Thus,$y e^{-3x} = -\frac{2}{3}x e^{-3x} - \frac{2}{9} e^{-3x} + c$.Multiplying by $e^{3x}$: $y = -\frac{2}{3}x - \frac{2}{9} + c e^{3x}$.Since the curve passes through $(1,2)$: $2 = -\frac{2}{3}(1) - \frac{2}{9} + c e^3 \implies 2 = -\frac{8}{9} + c e^3 \implies c e^3 = \frac{26}{9} \implies c = \frac{26}{9} e^{-3}$.Substituting $c$: $y = -\frac{2}{3}x - \frac{2}{9} + \frac{26}{9} e^{-3} e^{3x} \implies 9y = -6x - 2 + 26 e^{3x-3} \implies 6x + 9y + 2 = 26 e^{3x-3}$.

The equation of the curve passing through $(1,2)$ and whose tangent at any point $(x, y)$ makes an angle $\tan ^{-1}(2 x+3 y)$ with the $X$-axis is .........

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