How can i solve this differencial equation? : #y'+x^2 y=x^2#

Question

Emma Aldridge · Accepted Answer

# y = 1 + Ce^(-1/3x^3) #

When we have a First Order Linear non-homogeneous Ordinary Differential Equation of the following form, we can use an integrating factor; # dy/dx + P(x)y=Q(x) # We have: # y'+x^2y=x^2 # ..... [1] This is a First Order Ordinary Differential Equation in Standard Form. So we compute and integrating factor, #I#, using; # I = e^(int P(x) dx) # # \ \ = exp(int \ x^2 \ dx) # # \ \ = exp( 1/3x^3 ) # # \ \ = e^(1/3x^3) # And if we multiply the DE [1] by this Integrating Factor, #I#, we will have a perfect product differential; # y'e^(1/3x^3)+x^2e^(1/3x^3)y = x^2e^(1/3x^3) # # :. d/dx( ye^(1/3x^3) ) = x^2e^(1/3x^3) # Our original ODE has now become a Separable ODE as a result of which we can "separate the variables" to obtain:: # ye^(1/3x^3) = int \ x^2e^(1/3x^3) \ dx + C # It is easy to integrate this, and we obtain: # ye^(1/3x^3) = e^(1/3x^3) + C # Finally, arriving at the clear General Solution: # y = e^(-1/3x^3){e^(1/3x^3) + C} # # \ \ = 1 + Ce^(-1/3x^3) #

Emilia Aguilar · Answer

A simpler method to the one given in the other answer

#y'+x^2y=x^2# This ODE is separable. #dy/dx=x^2-x^2y=x^2(1-y)# #rArr int1/(1-y)dy=intx^2dx# #rArr-ln|1-y|=1/3x^3+"c"# #1-y=Be^(1/3x^3)# where #B=e^(-"c")# #y=1+Ae^(1/3x^3)# where #A=-B#

Emily Ammerman · Answer

undefined To solve the differential equation y' + x^2y = x^2, you can use an integrating factor approach. First, rearrange the equation into the standard form for linear first-order differential equations, which is y' + P(x)y = Q(x), where P(x) is the coefficient of y and Q(x) is the function on the right side of the equation. Then, determine the integrating factor, which is e^(∫P(x)dx). In this case, P(x) = x^2, so the integrating factor is e^(∫x^2 dx). Next, multiply both sides of the equation by the integrating factor. This gives you the left side as (e^(∫x^2 dx)) * y' + (e^(∫x^2 dx)) * x^2 * y. Now, recognize that the left side is the derivative of (e^(∫x^2 dx)) * y with respect to x. Integrate both sides of the equation with respect to x to find the general solution. Finally, you may need to apply initial conditions if they are provided to find the particular solution.