Calculus (Differntial Equation)

Question

Prove that this equation Y^-3 =x³(3e^x+C) is a solution of the given differential equation which is xy'+y+x⁴y⁴e^x=0

Answer 1

You want to take your solution and solve for y and y'

The plug those values in to the the differential equation.

Our solution is y^-3 = x³(3e^x+C)

Let's solve for y:

We'll do this in an interesting way here. Multiply both sides by y⁴

y = x³y⁴(3e^x+C)

Now solve for y' using implicit differentiation:

-3y^-4 * y' = x³(3e^x) + (3e^x +C)(3x²)

y^-4 * y' = -x³(e^x) - (3e^x +C)(x²)

y' = y⁴[-x³(e^x) - (3e^x +C)(x²)]

y' = -x²y⁴(xe^x +3e^x + C)

Now, plug y' into the differential equation

x [ -x²y⁴(xe^x +3e^x + C) ] + [x^-1(3e^x+C)^-1/3] + x⁴y⁴e^x= 0

Simplify, the equation:

-x⁴y⁴e^x -3x³y⁴e^x -Cx³y⁴ + y + x⁴y⁴e^x = 0

The first and last terms cancel

-3x³y⁴e^x -Cx³y⁴ + y = 0

Now plug what we got for y into just that last term

-3x³y⁴e^x -Cx³y⁴ + x³y⁴(3e^x+C) = 0

Simplify:

-x³y⁴(3e^x+C) + x³y⁴(3e^x+C) = 0

0 = 0

Therefore, when we plug in our solution, we get 0 = 0.

Therefore, we verified that y^-3 = x³(3e^x+C) is a solution

Answer 2

xy' + y+ y^4x^4e^x=0

divide by x:

y' + y/x= -y^4x^3e^x

divide by y^4

(y'/y^4) + (1/x(y^4))= -x^3e^x ... eq1

let z = 1/y^3

then, z'=-3y'/y^4

y'/y^4=-z'/3

substitute into eq 1:

-z'/3 + z/x = -x^3e^x ... eq 2: This is Bernoulli's differential equation.

solve by variation of parameters:

-z'/3 + z/x = 0

dz/z = 3dx/x

integrate to find solution z:

solution z = C1x^3 such that C1= C1(x) ... eq 3

z' = C1'x^3 + 3C1x^2

substitute into eq 2 to find C1:

then, (C1')X^3/3 = X^3e^x

dC1 = 3e^xdx

integrate to find C1:

C1 = 3e^x + C

substitute C1 into eq 3:

z = x^3 (3e^x +C)

since z = y^-3;

Then y^-3 = x^3(3e^x + C) is a general solution of this DE.