answer: -2\n# $\frac{dw}{dt}=100t(w + 2)$ $w(0)=7$\nparticular solution: \nto advance in the circuit, find…

Explanation:

Step1: Separate variables

Separate the variables in the differential equation (\frac{dW}{dt}=100t(W + 2)). We get (\frac{dW}{W + 2}=100t;dt).

Step2: Integrate both sides

Integrate (\int\frac{dW}{W + 2}=\int100t;dt). The left - hand side integral: (\int\frac{dW}{W + 2}=\ln|W + 2|+C_1). The right - hand side integral: (\int100t;dt=100\times\frac{t^{2}}{2}+C_2 = 50t^{2}+C_2). So, (\ln|W + 2|=50t^{2}+C) (where (C = C_2 - C_1)).

Step3: Use the initial condition (W(0)=7)

Substitute (t = 0) and (W=7) into (\ln|W + 2|=50t^{2}+C). (\ln|7 + 2|=50\times0^{2}+C), so (C=\ln(9)). Then (\ln|W + 2|=50t^{2}+\ln(9)). Using the property of logarithms (\ln a-\ln b=\ln\frac{a}{b}), we can rewrite it as (\ln\left(\frac{W + 2}{9}\right)=50t^{2}). Exponentiating both sides gives (\frac{W+2}{9}=e^{50t^{2}}), so (W = 9e^{50t^{2}}-2).

Step4: Find (W(\frac{1}{5}))

Substitute (t=\frac{1}{5}) into (W = 9e^{50t^{2}}-2). (t^{2}=\frac{1}{25}), and (50t^{2}=2). So (W(\frac{1}{5})=9e^{2}-2).

Answer:

(9e^{2}-2)