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Solve the following differential equation x2 dydx = x2 + xy − y2 - Mathematics and Statistics

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Question

Solve the following differential equation

`x^2  ("d"y)/("d"x)` = x2 + xy − y2 

Sum
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Solution

`x^2  ("d"y)/("d"x)` = x2 + xy − y2 

∴ `("d"y)/("d"x) = 1 + y/x - (y/x)^2`   .....(i)

Put `y/x` = t      .....(ii)

∴ y = tx

Differentiating w.r.t. x, we get

`("d"y)/("d"x) = "t" + x  ("dt")/("d"x)`   .....(iii)

Substituting (ii) and (iii) in (i), we get

`"t" + x  "dt"/("d"x)` = 1 + t − t2

∴ `x  "dt"/("d"x)` = 1 − t2

∴ `"dt"/(1 - "t"^2) = ("d"x)/x`

Integrating on both sides, we get

`int  "dt"/(1 - "t"^2) = int ("d"x)/x`

∴ `1/2 log|(1 + t)/(1 - t)|` = log |x| + log |c1|

∴ `log |(1 + y/x)/(1 - y/x)|` = 2log |x| + 2log |c1|

∴ `log|(x + y)/(x - y)|` = log |x2| + log |c12|

∴ `log|(x + y)/(x - y)|` = log |c1x2|

∴ `(x + y)/(x - y)` = c12x2

∴ `(x + y)/(x - y)` = cx2, where c = c12 

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Chapter 1.8: Differential Equation and Applications - Q.5

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SCERT Maharashtra Mathematics and Statistics (Commerce) [English] 12 Standard HSC
Chapter 1.8 Differential Equation and Applications
Q.5 | Q 10
SCERT Maharashtra Mathematics and Statistics (Arts and Science) [English] 12 Standard HSC
Chapter 2.6 Differential Equations
Attempt the following questions III | Q 10

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