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son4ous [18]
3 years ago
11

Find the value of kk for which the constant function x(t)=kx(t)=k is a solution of the differential equation 5t3dxdt+2x−2=05t3dx

dt+2x−2=0
Mathematics
1 answer:
uranmaximum [27]3 years ago
3 0
Given the differential equation

5t^3 \frac{dx}{dt} +2x-2=0

The solution is as follows:

5t^3 \frac{dx}{dt} +2x-2=0 \\  \\ \Rightarrow5t^3 \frac{dx}{dt} =2-2x \\  \\ \Rightarrow \frac{5}{2-2x} dx= \frac{1}{t^3} dt \\  \\ \Rightarrow \int {\frac{5}{2-2x} } \, dx = \int {\frac{1}{t^3}} \, dt \\  \\ \Rightarrow- \frac{5}{2} \ln(2-2x)=- \frac{1}{2t^2} +A \\  \\ \Rightarrow\ln(2-2x)= \frac{1}{5t^2} +B\\  \\ \Rightarrow2-2x=Ce^{\frac{1}{5t^2}} \\  \\ \Rightarrow 2x=Ce^{\frac{1}{5t^2}}+2 \\  \\ \Rightarrow x=De^{\frac{1}{5t^2}}+1
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