Find the particular solution of the differential equation...?

∫ e^2y dy = ∫ x - 5 dx

=> (1/2)e^2y = x²/2 - 5x + C

so, (1/2)e^2ln(5) = 5²/2 - 25 + C

=> 25/2 = 25/2 - 25 + C

i.e. C = 25

so, (1/2)e^2y = x²/2 - 5x + 25

=> e^2y = x² - 10x + 50

=> 2y = ln(x² - 10x + 50)

Therefore, y = (1/2)ln(x² - 10x + 50)

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Multiply through by e^(2y):

e^(2y) dy/dx = (x-5)

Multiply through by the differential operator dx:

∫ e^(2y) dy = ∫ (x-5) dx

(1/2) e^(2y) = (x^2)/2 - 5x + c

Multiply through by 2, recalling that we don't need to alter the arbitrary constant if it only becomes a different arbitrary constant:

e^(2y) = x^2 - 10x + c

y(5) = ln(5)

e^[2ln(5)] = 5^2 - 10(5) + c

e^ln(25) = 25 - 50 + c

25 = -25 + c

c = 50

e^(2y) = x^2 - 10x + 50

Take logs of both sides:

ln(e^[2y]) = ln(x^2 - 10x + 50)

2y = ln(x^2 - 10x + 50)

y = ln(x^2 - 10x + 50)/2

dy/dx = (x-5)(e^(-2y))

=> e^(2y) dy = (x-5) dx

=> (1/2)*e^(2y) dy = xdx - 5dx

=> (1/2)e^(2y) = x^2/2 - 5x + c

=> (1/2)e^(2ln(5)) = (5)^2/2 - 5*(5) + c

=> (1/2)*e^{ln(5)^2} = 25/2 - 25 + c

=> 5^2/2 = 25/2 - 25 + c

=> c = 25

Now plugging it in the solution with the 'c'


=> (1/2)e^(2y) = x^2/2 - 5x + 25

multiplying by 2 both sides ...............

=> e^(2y) = x^2 - 25x + 50

=> e^(2y) = e^{ln(x^2 - 25x + 50)}

=> 2y = ln(x^2 - 25x + 50)

=> y = (1/2)* ln(x^2 - 25x + 50) ======> Answer.

e^(2y) dy = (x-5) dx {Integrate both sides}

(e^(2y)) / 2 = x^2 / 2 - 5x + C {Take Ln of both sides}

2y = Ln[x^2 / 2 - 5x + C]

y = Ln[sqrt[x^2 / 2 - 5x + C]]

Now you need to sub in 5 for x to solve for C

Ln[sqrt[5^2 / 2 - 5*5 + C]] = Ln[5]

sqrt[5^2 / 2 - 5*5 + C] = 5

5^2 / 2 - 5*5 + C = 25

C = 25-12.5+25 = 37.5

Then y = Ln[x^2 / 2 - 5x + 37.5] / 2