Q. The particular solution of the differential equation dxdy−ex=yex, when x=0 and y=1 is
Rewrite Differential Equation: First, we need to rewrite the differential equation in a form that allows us to use the method of integrating factors. The equation is already in the form of a first-order linear differential equation, dxdy+P(x)y=Q(x), where P(x)=−ex and Q(x)=ex.
Calculate Integrating Factor: Next, we calculate the integrating factor, μ(x), which is e(∫P(x)dx). In this case, μ(x)=e(∫−exdx). The integral of −ex with respect to x is −ex, so μ(x)=e(−ex).
Multiply by Integrating Factor: Now, we multiply the entire differential equation by the integrating factor, μ(x), to get μ(x)⋅dxdy+μ(x)⋅P(x)⋅y=μ(x)⋅Q(x). Substituting the values of μ(x), P(x), and Q(x), we get e(−ex)⋅dxdy−e(−ex)⋅ex⋅y=e(−ex)⋅ex.
Recognize Derivative Form: We notice that the left side of the equation is now the derivative of (y⋅μ(x)) with respect to x. Therefore, we can write dxd(y⋅e(−ex))=e(−ex)⋅ex.
Integrate Both Sides: To solve for y, we integrate both sides of the equation with respect to x. The integral of the left side is y⋅e−ex, and the integral of the right side is ∫e−ex⋅exdx.
Re-evaluate Integrating Factor: The integral on the right side is not a standard integral and cannot be expressed in terms of elementary functions. It seems there has been a mistake in the calculation of the integrating factor or in the setup of the equation. We need to re-evaluate the integrating factor and the setup of the equation.
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