Question (1 point) Suppose that we use Euler's method to approximate the solution to the differential equation dy/dx=x^5/y; y(0.4)=7 Let . f(x,y)=x5/y. We let x0=0.4x0=0.4 and y0=7y0=7 and pick a step size h=0.2.h=0.2. Euler's method is the the following algorithm. From xnxn and ynyn, our approximations to the solution of the differential equation at the nth stage, we find the next stage by computing xn+1=xn+h,yn+1=yn+h⋅f(xn,yn).xn+1=xn+h,yn+1=yn+h⋅f(xn,yn). Complete the following table. Your answers should be accurate to at least seven decimal places. nn xnxn ynyn 00 0.40.4 77 11 22 33 44 55 The exact solution can also be found using separation of variables. It is y(x)=y(x)= Thus the actual value of the function at the point x=1.4x=1.4 y(1.4)=y(1.4)= .

OKGEZS The Asker · Calculus

(1 point) Suppose that we use Euler's method to approximate the solution to the differential equation

dy/dx=x^5/y; y(0.4)=7

Let . f(x,y)=x5/y.
We let x0=0.4x0=0.4 and y0=7y0=7 and pick a step size h=0.2.h=0.2. Euler's method is the the following algorithm. From xnxn and ynyn, our approximations to the solution of the differential equation at the nth stage, we find the next stage by computing

xn+1=xn+h,yn+1=yn+h⋅f(xn,yn).xn+1=xn+h,yn+1=yn+h⋅f(xn,yn).

Complete the following table. Your answers should be accurate to at least seven decimal places.

nn xnxn ynyn
00 0.40.4 77
11
22
33
44
55

The exact solution can also be found using separation of variables. It is
y(x)=y(x)=

Thus the actual value of the function at the point x=1.4x=1.4
y(1.4)=y(1.4)= .

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{:[(dy)/(dx)=(x^(5))/(y)","y(0:4)=7","h=0.2],[f(x","y)=(x^(5))/(y)","x_(0)=0.4","y_(0)=7","h=0.2],[x_(1)=x_(0)+h=0.6","x_(2)=x_(1)+h=0.8","x_(3)=x_(2)+h=1","x_(4)=x_(3)+h=1-2_(3)],[x_(5)=x_(4)+h=1.4]:}Euber's formula is given by y_(n+1)=y_(n)+hf(x_(n),y_(n)){:[n=0],[y_(1)=y_(0)+hf(x_(0),y_(0))=y_(0)+h[(x_(0)^(5))/(y_(0))]=7+(0.2)[((0.4)^(5))/(7)]]:}{:[=7+(0.2)[(0.01024)/(7)]=7+(0.2)(0.001462857143)],[y_(1)=7.000292571;*;],[y(0.6)=7.000292571],[y_(2)=y_(1)+hf(x_(1),y_(1))=y_(1)+h[(x_(1)^(5))/(y_(1))]=7.000292571+(0.2)[((0.6)^(5))/(7.00292571)]],[n=1],[=7.000292571+(0.2)[(0.07776)/(7.000292571)]],[=7.000292571+(0.2)[0.011108107]=7.000292571+0.002221621131],[y_(2)=7.002514192" i.e; "],[y(0.8)=7.002514192],[n=2],[y_(3)=y_(2)+hf(x_(2),y_(2))=y_(2)+h[(x_(2)5)/(y_(2))]=2.002514192+(0.2)[((0.8)^(5))/(7.002514192)]],[=7.002514192+(0.2)[(0.32768)/(7.002514192)]],[=7.002514192+(0.2)(0.0467946211=7.002514192+0.009358924267],[y_(3)=7.011873116" i.e; "],[y(1)=2.011873116]:}{:[n=3],[y_(4)=y_(3)+hf(x_(3),y_(3))=y_(3)+h[(x_(3)^(5))/(y_(3))]=7.011873116+(0.2)[((1)^(5))/(2.011873116 ... See the full answer