Solve 3y′+6xy=6e−x where y(0)=1.I am trying to solve 3y′+6xy=6e−x where y(0)=1. This is a first-order linear equation. Dividing by 3 and multiplying by ex2, we haveex2y′+2xex2y=2ex2−x.Thus, we have ddx(ex2y)=2ex2−x.Now, we have ex2y=2∫ex2−xdx.How do I calculate this integral? The solution is supposedly y(x)=(2x+1)e−x2.

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shvatismop1rj · Accepted Answer

Answer:Based on the reformulation of the given problem, we can proceed as follows in order to solve it:Therefore,3y′+6xy=6e−x2&amp;nbsp;⟺y′+2xy=2e−x2&amp;nbsp;&amp;nbsp;&amp;nbsp;⟺ex2y′+2xex2y=2&amp;nbsp;&amp;nbsp;&amp;nbsp;⟺(ex2y)′=2&amp;nbsp;&amp;nbsp;&amp;nbsp;⟺ex2y=2x+c&amp;nbsp;&amp;nbsp;&amp;nbsp;⟺y(x)=2xe−x2+ce−x2

Jadyn Gentry · Accepted Answer

Step 1The given solution was y(x)=(2x+1)e−x2. Supposing this is indeed the solution, we have3y′=−(12x2+6x−6)e−x2.Step 2Hence, 6xy=(12x2−6x)e−x2.Thus, we have 3y′+6xy=6e−x2, and indeed, the problem was misprinted. The problem should read &quot;Solve 3y′+6xy=6e−x2.&quot;

Solve \(\displaystyle{3}{y}'+{6}{x}{y}={6}{e}^{{-{x}}}\) where \(\displaystyle{y}{\left({0}\right)}={1}\). I am trying

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