## Calculus with Applications (10th Edition)

The actual result: $1.075$; The approximation: $1.094$; Difference: $1.094-1.075=0.019$
We are given $\frac{dy}{dx}=x^{2}y$ Using Euler's method $g(x,y)=x^{2}y$ Since $x=0, y=1$ $g(x_{0},y_{0})=0$ and $y_{1}=y_{0}+g(x_{0},y_{0})h=1+0\times0.1=1$ Now $x_{1}=0.1, y_{1}=1$ and $g(x_{1},y_{1})=0.01$ Then $y_{2}=1+(0.01)\times0.1=1.001$ $y_{3}=1.001+(0.04)\times0.1=1.005$ $y_{4}=1.005+(0.09)\times0.1=1.014$ $y_{5}=1.014+(0.162)\times0.1=1.0302$ $y_{6}=1.0302+(0.258)\times0.1=1.056$ $y_{7}=1.056+(0.38)\times0.1=1.095$ $y(0.6)=y_{7}=1.094$ $\frac{dy}{dx}=x^{2}y$ $\int \frac{1}{y}dy=\int x^{2} dx$ $\ln |y| =\frac{1}{3}x^{3}+C$ With x=0, y=1 $C=\ln |1| - \frac{1}{3}(0)^{3}=0$ $y(0.6) \rightarrow \ln|y|=\frac{1}{3}(0.6)^{3}=0.072$ $|y|=e^{0.072}=1.075$ Difference between the result and approximation: $1.094-1.075=0.019$