Answer to Question #158748 in Calculus for manish

Question

Solve the differential equation dy/dx = 2y+3e^x with x0 = 0, y0 = 0, using Taylor’s series method of order 2 to obtain the value of y at x = 0.1, 0.2.

Answer 1

$y'=2y+3e^x$

$\text {by the conditions of the problem, }$

$\text{the order of the Taylor series is 2}$

$y(x)\approx{y(x_0)}+\frac{y'(x_0)}{1!}(x-x_0)+\frac{y''(x_0)}{2!}(x-x_0)^2$

$\text {by the conditions of the problem }x_0=0 ;y_0=y(x_0)=0$

$y(x)\approx\frac{y'(0)}{1!}(x)+\frac{y''(0)}{2!}(x)^2$

$y'(0)=2y(0)+3e^0=3$

$y''=(y')'=2y'+(3e^x)'=2y'+3e^x$

$y''(0)=2y'(0)+3e^0=9$

$y(x)\approx3x+\frac{9}{2}x^2$

$y(0.1)= 3*0.1+4.5*0.1^2=0.345$

$y(0.2)= 3*0.2+4.5*0.2^2=0.78$

Answer:y(0.1)=0.345; y(0.2)=0.78