Find the solution of the differential equation that satisfies the given initial condition.?

x * ln(x) * dx = y * (1 + sqrt(15 + y^2)) * dy

u = 15 + y^2

du = 2y * dy

x * ln(x) * dx = (1/2) * (1 + sqrt(u)) * du

x * ln(x) * dx = (1/2) * du + (1/2) * u^(1/2) * du

a = ln(x)

da = dx/x

dv = x * dx

v = (1/2) * x^2

a * v - int(v * da) = (1/2) * u + (1/2) * (2/3) * u^(3/2) + C

(1/2) * x^2 * ln(x) - (1/2) * int(x^2 * dx/x) = (1/2) * u + (1/3) * u^(3/2) + C

x^2 * ln(x) - int(x * dx) = u + (2/3) * u^(3/2) + C

x^2 * ln(x) - (1/2) * x^2 = u + (2/3) * u^(3/2) + C

2x^2 * ln(x) - x^2 = 2u + (4/3) * u^(3/2) + C

6x^2 * ln(x) - 3x^2 = 6u + 4u^(3/2) + C

3x^2 * (2 * ln(x) - 1) = 2u * (3 + 2u^(1/2)) + C

3x^2 * (2 * ln(x) - 1) = 2 * (15 + y^2) * (3 + 2 * (15 + y^2)^(1/2)) + C

y = 1 when x = 1

3 * 1^2 * (2 * ln(1) - 1) = 2 * (15 + 1^2) * (3 + 2 * (15 + 1^2)^(1/2)) + C

3 * 1 * (2 * 0 - 1) = 2 * (15 + 1) * (3 + 2 * (15 + 1)^(1/2)) + C

3 * (-1) = 2 * 16 * (3 + 2 * 16^(1/2)) + C

-3 = 32 * (3 + 2 * 4) + C

-3 = 32 * (3 + 8) + C

-3 = 32 * 11 + C

-3 - 32 * 11 = C

-3 - 352 = C

-355 = C

3x^2 * (2 * ln(x) - 1) = 2 * (15 + y^2) * (3 + 2 * (15 + y^2)^(1/2)) - 355

2 * (15 + y^2) * (3 + 2 * (15 + y^2)^(1/2)) - 355 =>

2 * (15 + y^2) * 3 + 2 * 2 * (15 + y^2)^(3/2) - 355 =>

90 + 6y^2 + 4 * (15 + y^2)^(3/2) - 355 =>

6y^2 + 4 * (15 + y^2)^(3/2) - 265

3x^2 * (2 * ln(x) - 1) = 6y^2 + 4 * (15 + y^2)^(3/2) - 265

Here's the thing about computer-based entries.  If your answer doesn't match the answer key, or if the software is incapable of determining an equivalency, it will say it's wrong.

For instance, you're told to find the sum of 3 and 4.  The answer key would be 7.  But what if you answered with 2^2 + 2^1 + 2^0?  It's the same answer, but it doesn't match the key, so it's marked as wrong.  Unless the software is built to handle some calculations and isn't just doing a point-by-point comparison, you're gonna be screwed.

So if the solution you came up with results in the original question when you take a derivative, then it's the correct solution.

(1/2) * x^2 * ln(x) - (1/4) * x^2 + (355/3) = (1/2) * y^2 + (1/3) * (15 + y^2)^(3/2)

Let's make sure this passes through (1 , 1) first

(1/2) * 1 * ln(1) - (1/4) * 1 + 355/3 = (1/2) * 1 + (1/3) * (15 + 1)^(3/2)

(1/2) * 0 - (1/4) + 355/3 = (1/2) + (1/3) * 64

-1/4 + 355/3 = 1/2 + 64/3

355/3 - 64/3 = 1/2 + 1/4

291/3 = 3/4

97 = 3/4

So something is wrong here.

Let's derive what you have and see if we get the original differential equation

(1/2) * (x^2 * (1/x) + ln(x) * 2x) - (1/4) * 2x + 0 = (1/2) * 2y * y' + (1/3) * (3/2) * (15 + y^2)^(1/2) * 2y * y'

(1/2) * (x + 2x * ln(x)) - (1/2) * x = y * y' + (1/2) * 2y * y' * (15 + y^2)^(1/2)

(1/2) * (x + 2x * ln(x) - x) = y * y' * (1 + (1/2) * 2 * (15 + y^2)^(1/2))

(1/2) * 2x * ln(x) = y * y' * (1 + (15 + y^2)^(1/2))

x * ln(x) = y * y' * (1 + 15 + y^2)^(1/2))

That looks good.  You just made a mistake with your +C

(1/2) * x^2 * ln(x) - (1/4) * x^2 + (355/3) = (1/2) * y^2 + (1/3) * (15 + y^2)^(3/2)

Let's turn that into

(1/2) * x^2 * ln(x) - (1/4) * x^2 + C = (1/2) * y^2 + (1/3) * (15 + y^2)^(3/2)

Solve for c

(1/2) * 1^2 * ln(1) - (1/4) * 1^2 + C = (1/2) * 1^2 + (1/3) * (15 + 1^2)^(3/2)

(1/2) * 1 * 0 - (1/4) * 1 + C = (1/2) * 1 + (1/3) * 16^(3/2)

-1/4 + C = 1/2 + (1/3) * 64

C = 3/4 + 64/3

C = 9/12 + 256/12

C = 265/12

(1/2) * x^2 * ln(x) - (1/4) * x^2 + (265/12) = (1/2) * y^2 + (1/3) * (15 + y^2)^(3/2)

Try that one on for size.

This is direct integration. Rewrite the DE as x ln(x) dx =  (y + y*√(15+y²) )dy

The LHS can be done by parts. The RHS is direct.