Find the particular solution of the differential equation?

Question

dy/dx = (x+3)(e^-2y) satisfying the initial condition y(3) = ln(3). I tried separation but then I'm not sure how to integrate correctly and solve for the constant.

Answer 1

this is what i did

dy/dx=(x+3)e^-2y
dy/e^-2y = x+3 dx
e^2y dy = x+3 dx

integrate both sides
(1/2)e^2y = (1/2)x^2+3x+A
e^2y = x^2+6x+ 2A
raise both side by ln
2y= ln ( x^2+6x+ 2A )
y= 1/2 * ln ( x^2+6x+ 2A )
now, we know y(3) = ln(3) so substitute 3 into x and make the ean equal to ln 3

y(3) = 1/2 * ln(9+18+2A) = ln3

ln(9+18+2A)^1/2 = ln3
(9+18+2A)^1/2 = 3

9+18+2A = 9
18 + 2A = 0
2A = -18
A= -9

so y= 1/2 * ln ( x^2+6x-18 )

Answer 2

♠ I don’t think you tried it hard LOL: exp(2y)*dy =(x+3)*dx, hence
♣ 0.5*exp(2y) = 0.5*(x+3)^2 +0.5*C; 2y=ln((x+3)^2 +C); and
♦ y=0.5*ln((x+3)^2+C);
♥ now y(3) =ln(3); and ln(3) =0.5*ln((3+3)^2+C), hence 6^2+C=9, C=-27 and y(x)=0.5ln|(x+3)^2-27| is particular solution to compensate your family troubles;

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