Solution of first order integro-differential equationConsider dydx+p(x)y=∫0∞y(x)dx=R≈(say)y(0)=∫0∞f(x)y(x)dx.I want to solve above differential equation. Here is my try:y(x)=y(0)e−∫0xp(τ)dτ+R∫0xe−∫sxp(τ)dτds.where R=∫0∞y(0)e−∫0xp(τ)dτdx+R∫0∞∫0xe−∫sxp(τ)dτdsdxy(0)=∫0∞f(x)y(0)e−∫0xp(τ)dτdx+R∫0∞f(x)∫0xe−∫sxp(τ)dτdsdx.Now, I am not able to find R which is independent of y(0) and y(0) which is independent of R. Please let me know if it is possible to find solution of this integro-differential equation.

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Tatairfzk · Accepted Answer

Step 1Given:&amp;nbsp;R=y(0)∫0∞e-∫0xp(τ) dτ dx+R∫0∞∫0xe-∫sxp(τ) dτ ds dxy(0)=y(0)∫0∞f(x)e-∫0xp(τ) dτ dx+R∫0∞f(x)∫0xe-∫sxp(τ) dτ ds dxy(0)=y(0)∫0∞f(x)e-∫0xp(τ) dτ dx+R∫0∞f(x)∫0xe-∫sxp(τ) dτ ds dxyou can rearrange to obtain two linear equations with two unknowns. The former equation is equivalent to0=y(0)∫0∞e-∫0xp(τ) dτ dx+R∫0∞∫0xe-∫sxp(τ) dτ ds dx-1Step 2while the latter is equivalent to0=y(0)∫0∞f(x)e-∫0xp(τ) dτ dx-1+R∫0∞f(x)∫0xe-∫sxp(τ) dτ ds dxThis can be solved with just elementary algebra, or using matrix algebra. You will get y(0) and R exclusively in terms of f and p, provided suitable integrability conditions for these functions. It depends on the determinant of the corresponding matrix. If the matrix is invertible, then y(0) and R are unique.

Solution of first order integro-differential equation Consider \frac{dy}{dx} + p(x)y = \int_{

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