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

solution:

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

he sells x mount because thats eqivilqant to the maz and then divided into quarters

step-by-step explanation: