Find the solution of this differential equation whose graph it is through the point (1, 3e).Let y1(x)=ex and y2(x)=x2+1+ex be solutions of the differential equationa(x)y′+b(x)y=c(x)where a, b, c are continuous functions on R and the function a has no zeros. Find the solution of this differential equation whose graph it is through the point (1,3e).I know that difference between y1(x) and y2(x) is a solution of homogeneous differential equation. So solution of a homogeneous differential equation isyh=A(x2+1)I do not know how to get the particular solution. If I use variation constant for yp=z(x)(x2+1), where is z(x) a function, then I getz=∫c(x)a(x)e∫b(x)a(x)dxdx

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Find the solution of this differential equation whose graph it is through the point (1, 3e).Let y1(x)=ex and y2(x)=x2+1+ex be solutions of the differential equationa(x)y′+b(x)y=c(x)where a, b, c are continuous functions on R and the function a has no zeros. Find the solution of this differential equation whose graph it is through the point (1,3e).I know that difference between y1(x) and y2(x) is a solution of homogeneous differential equation. So solution of a homogeneous differential equation isyh=A(x2+1)I do not know how to get the particular solution. If I use variation constant for yp=z(x)(x2+1), where is z(x) a function, then I getz=∫c(x)a(x)e∫b(x)a(x)dxdx

Ouhamiptkg · Accepted Answer

Explanation:The general solution to a first-order DE is given by y(x)=yh(x)+yp(x)=cg(x)+yp(x), where c is arbitrary and g is a solution to the associated homogeneous equation. Since you know two solutions y1 and y2, the difference must be proportional to g, which means that g(x)=x2+1 as you predicted. Moreover, this implies that the particular solution is yp(x)=ex. Thus, we have y(x)=c(x2+1)+ex and you can now solve for the constant c using the initial condition y(1)=3e.

Find the solution of this differential equation whose

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