The value of y(0.9) using Euler's method for the given ODE for a step size of h=0.3 is approximately [math: 3dydx+5y2=sin⁡x,y(0.3)=5]

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Euler's update formula is y_{n+1} = y_n + h f(x_n,y_n) with f(x,y) = dy/dx = (sin x - 5y^2)/3. Starting at x0 = 0.3, y0 = 5 and using h = 0.3 we need two steps to reach x = 0.9: x = 0.3 → 0.6 → 0.9.

Step 1 (x0 = 0.3, y0 = 5):
- Compute f0 = (sin 0.3 - 5·5^2)/3 = (sin 0.3 - 125)/3.
- sin 0.3 ≈ 0.2955202067 so numerator = 0.2955202067 - 125 = -124.7044797933.
- f0 ≈ -124.7044797933 / 3 = -41.5681599311.
- Delta y = h·f0 = 0.3·(-41.5681599311) = -12.4704479793.
- y1 = y0 + Delta y = 5 - 12.4704479793 = -7.4704479793.

Step 2 (x1 = 0.6, y1 ≈ -7.4704479793):
- Compute y1^2 ≈ (-7.4704479793)^2 ≈ 55.807593 (rounded).
- Compute f1 = (sin 0.6 - 5·y1^2)/3 = (sin 0.6 - 5·55.807593)/3.
- sin 0.6 ≈ 0.5646424734 and 5·y1^2 ≈ 279.037965, so numerator ≈ 0.5646424734 - 279.037965 = -278.4733225266.
- f1 ≈ -278.4733225266 / 3 = -92.8244408422.
- Delta y = h·f1 = 0.3·(-92.8244408422) = -27.8473322527.
- y2 = y1 + Delta y ≈ -7.4704479793 - 27.8473322527 = -35.3177802320.

Thus Euler's approximation y(0.9) ≈ -35.31778, which rounds to -35.318 (option d). The other choices (-58.918, -36.458, -69.058) do not match the numerical two-step Euler computation above; option b (-36.458) is close but differs by about 1.14, so it is not the correct Euler result. Confirmed final answer: d. -35.318.

The value of y(0.9) using Euler's method for the given ODE for a step size of h=0.3 is approximately [math: 3dydx+5y2=sin⁡x,y(0.3)=5]Single choice

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