<p><strong>Find the equation of a curve passing through the origin if the slope of the tangent to the curve at any point <math><mrow><mrow><mo>(</mo><mrow><mi>x</mi><mo>,</mo><mi>y</mi></mrow><mo>)</mo></mrow></mrow></math> is equal to the square of the difference of the abscissa and ordinate of the point.</strong></p>

This is a Long Answer Type Questions as classified in NCERT ExemplarSol:Here, slope of the of the curve=dydxand the difference between the abscissa and ordinate =x−y.∴ As per the condition, dydx=(x−y)2Put x−y=v 1−dydx=dvdx∴ dydx=1−dvdx∴ the equation becomes 1−dvdx=v2 ⇒dvdx=1−v2 ⇒dv1−v2=dxIntegrating both sides, we get ∫dv1−v2=∫dx⇒12log|1+v1−v|=x+c ⇒12log|1+x−y1−x+y|=x+c …(1)Since, the curve is through (0,0), then⇒12log|1+0−01−0+0|=0+c ⇒c=0∴ On putting c=0 in eq.(1) we get 12log|1+x−y1−x+y|=x ⇒log|1+x−y1−x+y|=2x∴ 1+x−y1−x+y=e2x⇒ (1+x−y)=e2x(1−x+y)Hence, the required equation is (1+x−y)=e2x(1−x+y).

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Ncert Solutions Maths class 12th Maths Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Nine Differential Equations

Find the equation of a curve passing through the origin if the slope of the tangent to the curve at any point $(x, y)$ is equal to the square of the difference of the abscissa and ordinate of the point.

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alok kumar singh | Contributor-Level 10

3 months ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

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$\begin{array}{l} H e r e, s l o p e o f t h e o f t h e c u r v e = \frac{d y}{d x} \\ a n d t h e d i f f e r e n c e b e t w e e n t h e a b s c i s s a a n d o r d i n a t e = x - y . \\ ∴ A s p e r t h e c o n d i t i o n, \frac{d y}{d x} = {(x - y)}^{2} \\ P u t x - y = v \\ 1 - \frac{d y}{d x} = \frac{d v}{d x} \\ ∴ \frac{d y}{d x} = 1 - \frac{d v}{d x} \\ ∴ t h e e q u a t i o n b e c o m e s \\ 1 - \frac{d v}{d x} = v^{2} \Rightarrow \frac{d v}{d x} = 1 - v^{2} \Rightarrow \frac{d v}{1 - v^{2}} = d x \\ I n t e g r a t i n g b o t h s i d e s, w e g e t \\ \int \frac{d v}{1 - v^{2}} = \int d x \\ \Rightarrow \frac{1}{2} l o g | \frac{1 + v}{1 - v} | = x + c \Rightarrow \frac{1}{2} l o g | \frac{1 + x - y}{1 - x + y} | = x + c \dots (1) \\ Since, t h e c u r v e i s t h r o u g h (0, 0), t h e n \\ \Rightarrow \frac{1}{2} l o g | \frac{1 + 0 - 0}{1 - 0 + 0} | = 0 + c \Rightarrow c = 0 \\ ∴ O n p u t t i n g c = 0 i n e q . (1) w e g e t \\ \frac{1}{2} l o g | \frac{1 + x - y}{1 - x + y} | = x \Rightarrow l o g | \frac{1 + x - y}{1 - x + y} | = 2 x \\ ∴ \frac{1 + x - y}{1 - x + y} = e^{2 x} \\ \Rightarrow (1 + x - y) = e^{2 x} (1 - x + y) \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s (1 + x - y) = e^{2 x} (1 - x + y) . \end{array}$

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Find the equation of a curve passing through the origin if the slope of the tangent to the curve at any point $(x, y)$ is equal to the square of the difference of the abscissa and ordinate of the point.

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Find the equation of a curve passing through the origin if the slope of the tangent to the curve at any point $(x, y)$ is equal to the square of the difference of the abscissa and ordinate of the point.