How do you solve the differential equation #(dy)/dx=e^(y-x)sec(y)(1+x^2)#, where #y(0)=0# ?

Question

Cameron Alexander · Accepted Answer

By separating variables and integrating, #int e^{-y}cosy dy=int (1+x^2)e^{-x} dx# Integration by Parts 1
#u=e^{-y}" "dv=cosy dy#.
#du=-e^{-y} dy"  "v=siny#. Integration by Parts 2
#u=e^{-y}" "dv=siny dy#.
#du=-e^{-y}dy" "v=-cosy# By Integration by Pats 1, #(LHS)=e^{-y}siny+int e^{-y}siny dy# by Integration by Parts 2, #=e^{-y}siny-e^{-y}cosy-int e^{-y}cosy dy# Since the last integral is the same as #(LHS)#, we have #(LHS)=e^{-y}(siny-cosy)-(LHS)# by adding #(LHS)#, #Rightarrow2(LHS)=e^{-y}(siny-cosy)# by dividing by #2#, #Rightarrow (LHS)=e^{-y}/2(siny-cosy)# Integration by Parts 3
#u=1+x^2" "dv=e^{-x}dx#
#du=2xdx" "v=-e^{-x}# Integration by Parts 4
#u=2x" "dv=e^{-x}dx#
#du=2dx" "v=-e^{-x}#  Let us evaluate the right-hand side. By Integration by Parts 3, #(RHS)=-(1+x^2)e^{-x}+int2xe^{-x}dx# by Integration by Parts 4, #=-(1+x^2)-2xe^{-x}+int 2e^{-x} dx# #=-(x^2+2x+1)e^{-x}-2e^{-x}+C# #=-(x^2+2x+3)e^{-x}+C# By setting #(LHS)=(RHS)#, #e^{-y}/2(siny-cosy)=-(x^2+2x+3)e^{-x}+C# Since #y(0)=0#, we have #Rightarrow 1/2(0-1)=-3+C Rightarrow C=5/2# Hence, the solution is implicitly expressed as #e^{-y}/2(siny-cosy)=-(x^2+2x+3)e^{-x}+5/2#. I hope that this was helpful.

Emily Ammerman · Answer

undefined To solve the differential equation $ \frac{dy}{dx} = e^{y-x}\sec(y)(1+x^2) $ with the initial condition $ y(0) = 0 $, follow these steps:

1. Separate the variables.
2. Integrate both sides.
3. Solve for the constant of integration using the initial condition.

Step 1: Separate the variables:
   $ \frac{dy}{dx} = e^{y-x}\sec(y)(1+x^2) $
   $ e^{-y}\sec(y) \, dy = (1+x^2) \, dx $

Step 2: Integrate both sides:
   $ \int e^{-y}\sec(y) \, dy = \int (1+x^2) \, dx $

Step 3: Solve for the constant of integration using the initial condition:
   $ y(0) = 0 $

After solving the integrals and adding the constant of integration, we will have a general solution containing an arbitrary constant. Then, we'll use the initial condition to determine the value of the constant.

Unfortunately, the integral of $ e^{-y}\sec(y) \, dy $ does not have a simple closed form, making this problem quite challenging to solve analytically. It may require numerical methods or approximation techniques to find a solution.

Therefore, the final solution involves solving the integral and applying the initial condition to determine the constant of integration, but the integral itself cannot be expressed in elementary functions.