What is a solution to the differential equation #yy'=e^x=0# with y=4 when x=0?

Related questions

• How do you solve separable differential equations?
• How do you solve separable first-order differential equations?
• How do you solve separable differential equations with initial conditions?
• What are separable differential equations?
• How do you solve the differential equation #dy/dx=6y^2x#, where #y(1)=1/25# ?
• How do you solve the differential equation #y'=e^(-y)(2x-4)#, where #y5)=0# ?
• How do you solve the differential equation #(dy)/dx=e^(y-x)sec(y)(1+x^2)#, where #y(0)=0# ?
• How do I solve the equation #dy/dt = 2y - 10#?
• Given the general solution to #t^2y'' - 4ty' + 4y = 0# is #y= c_1t + c_2t^4#, how do I solve the...
• How do I solve the differential equation #xy'-y=3xy, y_1=0#?