from a handpicked tutor in LIVE 1-to-1 classes

# Determine by inspection two solutions of the given first-order IVP. y' = 4y^{3/}4, y(0) = 0

**Solution:**

Given, y' = 4y^{3/4}, 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' = 4y^{3/4}

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' = 4y^{3/4}

x’ = 4x^{3/4}

1 = 4x^{3/4} not balanced equation.

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

Now, consider y = x^{3} passing through the origin

Substitute y = x^{3} in y' = 4y^{3/4}

x^{3} = 4(x^{3})^{3/4}

3x^{2} = 4x^{9/4} not balanced equation.

Therefore, y = x^{3} is not a solution curve of the given differential equation.

Now, consider y = x^{4} passing through the origin

Substitute y = x^{4} in y' = 4y^{3/4}

x^{4} = 4(x^{4})^{3/4}

4x^{3} = 4x^{12/4}

4x^{3} = 4x^{3}

Therefore, y = x^{3} is a solution curve of the given differential equation.

So, the two solutions are y = 0 and y = x^{4}

Therefore, the two solutions are y = 0 and y = x^{4}.

## Determine by inspection two solutions of the given first-order IVP. y' = 4y^{3/4}, y(0) = 0

**Summary:**

By inspection two solutions of the given first-order ivp y' = 4y^{3/4}, y(0) = 0 are y = 0 and y = x^{4}.

visual curriculum