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Jazlinmoreno5683
13.07.2019 •
Mathematics
Solve the following initial-value problem, showing all work, including a clear general solution as well as the particular solution requested (label the gen. sol. and particular write both solutions in explicit form (solved for y). x dy/dx = x^3 + 2y subject to: y(2) = 6
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Ответ:
General Solution is
and the particular solution is ![y=x^{3}-\frac{1}{2}x^{2}](/tpl/images/0086/1647/f6e85.png)
Step-by-step explanation:
This is a linear diffrential equation of type
here![p(x)=\frac{-2}{x}](/tpl/images/0086/1647/016d9.png)
The solution of equation i is given by
we have![e^{\int p(x)dx}=e^{\int \frac{-2}{x}dx}\\\\e^{\int \frac{-2}{x}dx}=e^{-2ln(x)}\\\\=e^{ln(x^{-2})}\\\\=\frac{1}{x^{2} } \\\\\because e^{ln(f(x))}=f(x)]\\\\Thus\\\\e^{\int p(x)dx}=\frac{1}{x^{2}}](/tpl/images/0086/1647/c6dc4.png)
Thus the solution becomes
This is the general solution now to find the particular solution we put value of x=2 for which y=6
we have![6=8+4c](/tpl/images/0086/1647/9aa7b.png)
Thus solving for c we get c = -1/2
Thus particular solution becomes
Ответ:
it is already in scientific notation