# Dy/dx = x^3-y^3 Eulers method

• MHB
• karush
In summary, the conversation discussed using Euler's method to approximate the value of $y$ at $x=1$ on the solution curve to the differential equation $\dfrac{dy}{dx}=x^3-y^3$ that passes through $(0,0)$. It was suggested to use $\Delta x = \dfrac{1}{5}$ or 5 steps, and the resulting approximation of $y(1)$ was approximately 0.1599.

#### karush

Gold Member
MHB
Use Euler's method to approximate the value of y at $x=1$ on the solution curve to the differintial equation
$$\dfrac{dy}{dx}=x^3-y^3$$
that passes through $(0,0)$, Use $\Delta x = \dfrac{1}{5}$ or 5 steps

$\quad x_{1}=x_{0}+h=0+\frac{1}{5}=\frac{1}{5}$
$y\left(x_{1}\right)=y\left( \frac{1}{5} \right)=y_{1}=y_{0}+h \cdot f \left(x_{0}, y_{0} \right)=0+h \cdot f \left(0, 0 \right)=0 + \frac{1}{5} \cdot \left(0 \right)=0$
$\quad x_{2}=x_{1}+h=\frac{1}{5}+\frac{1}{5}=\frac{2}{5}$
$y\left(x_{2}\right)=y\left( \frac{2}{5} \right)=y_{2}=y_{1}+h \cdot f \left(x_{1}, y_{1} \right)=0+h \cdot f \left(\frac{1}{5}, 0 \right)=0 + \frac{1}{5} \cdot \left(0.008 \right)=0.0016$
$\quad x_{3}=x_{2}+h=\frac{2}{5}+\frac{1}{5}=\frac{3}{5}$
$y\left(x_{3}\right)=y\left( \frac{3}{5} \right)=y_{3}=y_{2}+h \cdot f \left(x_{2}, y_{2} \right)=0.0016+h \cdot f \left(\frac{2}{5}, 0.0016 \right)=0.0016 + \frac{1}{5} \cdot \left(0.0634 \right)\approx 0.0144$
$\quad x_{4}=x_{3}+h=\frac{3}{5}+\frac{1}{5}=\frac{4}{5}$
$y\left(x_{4}\right)=y\left( \frac{4}{5} \right)=y_{4}=y_{3}+h \cdot f \left(x_{3}, y_{3} \right)=0.01434+h \cdot f \left(\frac{3}{5}, 0.0144 \right)=0.01434 + \frac{1}{5} \cdot \left(0.2159 \right)\approx 0.0576$
$\quad x_{5}=x_{4}+h=\frac{4}{5}+\frac{1}{5}=1$
$y\left(x_{5}\right)=y\left( 1 \right)=y_{5}=y_{4}+h \cdot f \left(x_{4}, y_{4} \right)=0.0575+h \cdot f \left(\frac{4}{5}, 0.0576 \right)=0.0576 + \frac{1}{5} \cdot \left(0.5119 \right)=0.1599$
so
$y\left(1\right)\approx 0.1599$

ok I think this is sort of it thot there was some way to trim it down

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