Determine by inspection two solutions of the given first-order IVP. y' = 4y3/4, y(0) = 0
Solution:
Given, y' = 4y3/4, y(0) = 0
The initial condition y(0) = 0 implies 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' = 4y3/4
y’ = 0
Therefore, y = 0 is a solution curve of the given differential equation.
Consider the curve y = x, passing through the origin
Substitute y = x in y' = 4y3/4
x’ = 4x3/4
1 = 4x3/4 not balanced equation.
Therefore, y = x is not a solution curve of the differential equation.
Now, consider y = x3 passing through the origin
Substitute y = x3 in y' = 4y3/4
x3 = 4(x3)3/4
3x2 = 4x9/4 not balanced 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' = 4y3/4
x4 = 4(x4)3/4
4x3 = 4x12/4
4x3 = 4x3
Therefore, y = x3 is a solution curve of the given differential equation.
So, the two solutions are y = 0 and y = x4
Therefore, the two solutions are y = 0 and y = x4.
Determine by inspection two solutions of the given first-order IVP. y' = 4y3/4, y(0) = 0
Summary:
By inspection two solutions of the given first-order ivp y' = 4y3/4, y(0) = 0 are y = 0 and y = x4.
visual curriculum