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# Solve the differential equation $\begin{bmatrix}\large\frac{e^{-2\sqrt{x}}}{\sqrt{x}}-\large\frac{y}{\sqrt{x}}\end{bmatrix}\large\frac{dx}{dy}$$=1(x\neq0) Can you answer this question? ## 1 Answer 0 votes Toolbox: • To solve the first order linear differential equation: • (i) First write the equation of the form \large\frac{dy}{dx}$$ + py = Q$
• (ii) Find the integral factor(I.F) =$e^{\int pdx}$
• (iii) Write the solution as y.(I.F) - integration of Q.(I.F) + C
Step 1:
Given : $\big[\large\frac{e^{-2\sqrt x}}{\sqrt x}-\large\frac{y}{\sqrt x}\big]\large\frac{dx}{dy}$$=1 \large\frac{dx}{dy}=\frac{1}{\big[\Large\frac{e^{-2\sqrt x}}{\sqrt x}-\frac{y}{\sqrt x}\big]} \large\frac{dy}{dx}=\big[\Large\frac{e^{-2\sqrt x}}{\sqrt x}-\frac{y}{\sqrt x}\big] \large\frac{dy}{dx}+\frac{y}{\sqrt x}=\frac{e^{-2\sqrt x}}{\sqrt x} Here p=\large\frac{1}{\sqrt x} Q=\Large\frac{e^{-2\sqrt x}}{\sqrt x} Step 2: To find the integrating factor (I.F) integrate p. \int \large\frac{1.dx}{\sqrt x}=$$2\sqrt x$
Hence $I.F=e^{2\sqrt x}$
Step 3: