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
