Ex 9.5, 16 - Chapter 9 Class 12 Differential Equations

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Ex 9.5, 16 Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (𝑥 , 𝑦) is equal to the sum of the coordinates of the point.
We know that
Slope of tangent to curve at (x, y) = 𝑑𝑦/𝑑𝑥
Given that
Slope of the tangent to the curve at any point (𝑥 , 𝑦) is equal to the sum of the coordinates of the point.
Therefore,
𝑑𝑦/𝑑𝑥 = x + y
𝑑𝑦/𝑑𝑥 − y = x
This is the form 𝑑𝑦/𝑑𝑥+Py=Q
where P = −1 & Q = x
Finding Integrating factor
IF = 𝑒^∫1▒〖𝑝𝑑𝑥 〗
IF = 𝑒^∫1▒〖(−1)𝑑𝑥 〗
IF = e−x
Solution is
y(IF) = ∫1▒〖(𝑄×𝐼𝐹)𝑑𝑥+𝑐〗
𝑦𝑒^(−𝑥) = ∫1▒〖𝑥𝑒^(−𝑥) 𝑑𝑥+𝑐〗
ye−x = x ∫1▒〖𝑒^(−𝑥) 𝑑𝑥−〗 ∫1▒〖[1∫1▒〖𝑒^(−𝑥) 𝑑𝑥〗] 𝑑𝑥+𝑐〗
ye−x = −x 𝑒^(−𝑥) −∫1▒〖−𝑒^(−𝑥) 𝑑𝑥+𝑐〗
ye−x = −x 𝑒^(−𝑥)+∫1▒〖𝑒^(−𝑥) 𝑑𝑥+𝑐〗
ye−x = −x 𝑒^(−𝑥)+(−𝑒^(−𝑥))/(−1) +𝑐
ye−x = −x 𝑒^(−𝑥)−𝑒^(−𝑥) +𝑐
Dividing both sides by e−x
y = −x − 1 + 𝑐/𝑒^(−𝑥)
Integrating by parts with
∫1▒〖𝑓(𝑥) 𝑔(𝑥) 𝑑𝑥=𝑓(𝑥) ∫1▒〖𝑔(𝑥) 𝑑𝑥 −∫1▒〖[𝑓^′ (𝑥) ∫1▒〖𝑔(𝑥) 𝑑𝑥] 𝑑𝑥〗〗〗〗
Take f (x) = x & g (x) = 𝑒^(−𝑥)
y = −x − 1 + cex
Since curve passes through origin,
Putting x = 0 & y = 0 in (2)
0 = 0 − 1 + Ce0
1 = C
C = 1
Put value of C in (1)
y = −x − 1 + ex
x + y + 1 = ex
…(1)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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