Chapter 5: Applications of Integration
Section 5.6: Differential Equations
Graph the solution of the initial-value problem consisting of the differential equation y′=4⁢x⁢y2+8⁢y2+x+2, and the initial condition y1=0.
The right-hand side of the differential equation factors to 2+x1+4 y2, so the equation is separable. The solution of the IVP is
∫0y11+4 s2 ⅆs
Figure 5.6.4(a) Solution of IVP
The solution of the initial-value problem is contained in Figure 5.6.4(a). Only the branch of the tangent function that passes through 1,0 can be considered the solution of the IVP; all the other branches satisfy the differential equation but not the initial condition y1=0.
Solution via Context Panel
Control-drag the differential equation.
Press the Enter key.
Add an Initial Condition≻y1=0
(enter into the pop-up dialog)
Context Panel: Solve DE≻yx
→add initial condition
Factor the right-hand side of the differential equation
Control-drag the right-hand side of the differential equation.
Context Panel: Factor
4 x y2+8⁢y2+x+2= factor 4⁢y2+1⁢x+2
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