Q. et x(t) be the solution of the initial value problem:=−x−x3,x(0)=0,x′(0)=2. Vhen x=1, the value of 4(x˙)2 is ormula: x¨=dxd(2(x˙)2)
Given Differential Equation: Given the differential equation: x^{\¨} = -x - x^{3} and initial conditions x(0)=0, x′(0)=2.
Differentiate Velocity: To find x^{\¨}, we differentiate the velocity x^{\˙} with respect to x using the chain rule: x^{\¨} = \frac{d}{dx}\left(\frac{(x^{\˙})^2}{2}\right).
Set Equation Equal: We know that x^{\¨} = -x - x^{3}. So, we set (d)/(dx)((x^{\˙})^2/2) = -x - x^{3}.
Integrate Both Sides: Multiply both sides by 2 to get rid of the fraction: dxd(x˙)2 = −2x−2x3.
Find Integration Constant: We need to find (x˙)2 when x=1. Integrate both sides with respect to x to find (x˙)2 as a function of x.
Use Initial Condition: The integral of the right side with respect to x is −x2−21x4+C, where C is the integration constant.
Calculate Integration Constant: To find C, use the initial condition x′(0)=2, which means (x˙)2=4 when x=0.
Find Velocity Squared: Plug x=0 into −x2−21x4+C to get C: 4=−02−2104+C, so C=4.
Calculate Velocity Squared: Now we have (x˙)2=−x2−21x4+4. Plug in x=1 to find (x˙)2: (x˙)2=−12−2114+4.
Multiply Velocity Squared: Calculate (x(˙))2 when x=1: (x(˙))2=−1−(21)(1)+4.
Multiply Velocity Squared: Calculate (x(˙))2 when x=1: (x(˙))2=−1−(21)(1)+4. Simplify the expression: (x(˙))2=−1−21+4=2.5.
Multiply Velocity Squared: Calculate (x(˙))2 when x=1: (x(˙))2=−1−(21)(1)+4. Simplify the expression: (x(˙))2=−1−21+4=2.5. Finally, multiply (x(˙))2 by 4 to find 4(x(˙))2: 4(x(˙))2=4×2.5=10.
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