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HallsofIvy
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You might recognize it better if you subtract 0.5y from both sides to get
\(\displaystyle y'- 0.5y= 4e^{0.8x}\) with initial condition y(0)= 2.
\(\displaystyle y'- 0.5y= 4e^{0.8x}\) with initial condition y(0)= 2.
HallsofIvy
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This is a linear equation. It can be solved be separating into "homogeneous" and "non-homogeneous" parts as Rumsek suggests.
You can also find an "integrating factor, a function, h(x), such that multiplying by it, to get
\(\displaystyle h(x)y'+ 5h(x)y= 4h(x)e^{.8x}\) has an "exact derivative", so that d(hy)/dx= h dy/dx+ 5hy, on the left.
From the product rule, d(hy)/dx= h dy/dx+ (dh/dx)y= h dy/dx+ 5h y so we must have dh/dx= 5x.
You can also find an "integrating factor, a function, h(x), such that multiplying by it, to get
\(\displaystyle h(x)y'+ 5h(x)y= 4h(x)e^{.8x}\) has an "exact derivative", so that d(hy)/dx= h dy/dx+ 5hy, on the left.
From the product rule, d(hy)/dx= h dy/dx+ (dh/dx)y= h dy/dx+ 5h y so we must have dh/dx= 5x.