Question textGiven the IVP $y'-y^2-1=0$, $y(0)=0$ we have that $\displaystyle y(\frac{\pi}{4})=$ Answer 1 Question 7[input].

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We start from the given IVP: y' - y^2 - 1 = 0 with y(0) = 0.

First, rewrite the differential equation in a separable form: dy/dx = y^2 + 1.
Separate variables: dy/(y^2 + 1) = dx.
Integrate both sides: ∫ dy/(y^2 + 1) = ∫ dx, which gives arctan(y) = x + C.
Apply the initial condition y(0) = 0: arctan(0) = 0 = 0 + C, so C = 0.
Hence, arctan(y) = x. To solve for y as a function of x, take the tangent of both sides: y = tan(x).
Now evaluate at x = π/4: y(π/4) = tan(π/4).
Recall that tan(π/4) = 1, so the value is 1.

Key points to verify along the way:
- The equation is separable, allowing dy/(y^2+1) = dx.
- The antiderivative of 1/(y^2+1) is arctan(y), valid for all real y.
- The initial condition pins down the constant of integration to zero, yielding arctan(y) = x.
- Evaluating the tangent at π/4 yields 1, consistent with the standard special angle value.

If there are no answer choices provided, this is the method and calculation you would use to determine the value of y at x = π/4.

ENG1090 - MUM S1 2025 Lecture quiz 11

Question textGiven the IVP \(y'-y^2-1=0\), \(y(0)=0\) we have that \(\displaystyle y(\frac{\pi}{4})=\) Answer 1 Question 7[input].

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