Question

Determine by inspection two solutions of the given first-order ivp. y' = 5y4/5, y(0) = 0

Answer 1

The initial value problem y' = 5y^(4/5), y(0) = 0 has two solutions by inspection: y(t) = 0 and y(t) = t^5, where the constant in the general solution is undetermined due to the initial condition.

Step-by-step explanation:

Finding Solutions to a Differential Equation

The initial value problem (IVP) specified is y' = 5y4/5, y(0) = 0. By inspection, we are looking for functions whose derivative matches the expression on the right-hand side when raised to the 4/5 power and multiplied by 5. Two functions that satisfy this condition are y(t) = 0 and y(t) = t5. The first solution is trivial because the derivative of zero is also zero, which satisfies the differential equation and the initial condition. The second solution results from the general solution to this differential equation, which can be obtained by separating variables and integrating both sides, that occurs to be Ct5, where C is the constant of integration. Using the initial condition y(0) = 0 to solve for C, we find that C can be zero, which leads back to our first trivial solution, or C can be left undefined as the initial condition does not provide information to determine it. Therefore, t5 satisfies the differential equation for all values of C, making it a non-trivial solution.

Answer 2

Consider the equation,

Y=5y4/5, y(0) =0

The initial condition y(0) =0 given that the graph of the solution curves will pass through the origin.

Consider the curve y = 0,

Passing through the origin

Substitute y= 0 in y’ = 5y4/5

0=0 balenced the equation

Therefore y = o is a solution curve of the given differential equation

Consider the curve y = x,

Passing through the origin.

Substitute y = x in y’ = 5y4/5

(x’) = 5x4/5

1 = 5x4/5 not balanced the equation

Therefore, y = x is not a solution curve of the given differential equation.

Now consider y = x³ passing through the origin

Substitute y = x³ in y’=5y4/5

(X3) = 5(X³) 4/5

3x² = 5x12/5 not balanced the equation

Therefore y = X³ is not a solution curve of the given differential equation

Now consider y = X⁴ passing through the origin

Substitute y = X⁴ in y’ = 5y4/5

(X⁴)’ = 5(X⁴)4/5

4x³ = 5x16/5 not balanced the equation

Therefore y = X⁴ is not a solution curve of the given equation,

Now, y = x⁵ passing through the origin

Substitute y = x⁵

Y’ = 5y4/5

(X⁵) = 5(x5)4/5

5x⁴=5x⁴ balanced the equation

Therefore y= x⁵ is a solution curve of the given differential equation

Therefore, the two solution are y =0

Y = x⁵

The constant solution is y = 0

And the polynomial solution is y = x⁵