Respuesta :
Answer:
[tex]y = - [ \frac{1}{t - \frac{28}{9} } + t ][/tex]
Step-by-step explanation:
Solution:-
- A change of variable is a technique employed in solving many differential equations that are of the form: y ' = f ( t , y ).
- Considering a differential equation of the form y' = f ( αt + βy + γ ), where α, β, and γ are constants. A substitution of an arbitrary variable z = αt + βy + γ is made and the given differential equation is converted into a form: z ' = g ( z ).
- This substitution basically allow us to solve in-separable differential equations by converting them into a form that can be separated, followed by the set procedure.
- We are to solve the initial value problem for the following differential equation:
[tex]y' = ( t + y ) ^2 - 1 , y ( 3 ) = 6[/tex]
First Step: Make the appropriate substitution
- We will use a arbitrary variable ( z ) and define the our substitution by finding a multi-variable function f ( t , y ) that is a part of the given ODE.
- We see that the term ( t + y ) is a multi-variable function and also the culprit that doesn't allow us to separate our variables.
- Usually, the change of variable substitution is made for such " culprits ".
- So our substitution would be:
[tex]z = t + y[/tex]
Second Step: Implicit differential of the substitution variable ( z ) with respect to the independent variable
- In the given ODE we see that the variable ( t ) is our independent variable. So we will derivate the supposed substitution as follows:
[tex]\frac{dz}{dt} = 1 + \frac{dy}{dt} \\\\\frac{dy}{dt} = -1 + \frac{dz}{dt}[/tex]
Remember: z is a multivariable function of "t" and "y". So we perform implicit differential for the variable " z ".
Third Step: Plug in the differential form in step 2 and change of variable substitution of ( z ) in the given ODE.
- The given ODE can be expressed as:
[tex]\frac{dy}{dt} = ( t + y ) ^2 - 1\\\\\frac{dz}{dt} - 1 = ( z ) ^2 - 1\\\\\frac{dz}{dt} = z ^2 \\[/tex] ... Separable ODE
Fourth Step: Separate the variables and solve the ODE.
- We see that the substitution left us with a simple separable ODE.
Note: If we do not arrive at a separable ODE, then we must go back and re-choose our change of variable substitution for ( z ).
- We will progress by solving our ODE:
[tex]\frac{dz}{z^2} = dt\\\\\int {\frac{1}{z^2} } \, dz = \int {1} \, dt\\\\-\frac{1}{z} = t + c\\\\\frac{1}{z} = - (t + c )\\\\z = -\frac{1}{t + c}[/tex]
Where,
c: The constant of integration
Fifth Step: Back-substitution of variable ( z )
- We will now back-substitute the substitution made in the first step and arrive back at our original variables ( y and t ) as follows:
[tex]t + y = - \frac{1}{t + c} \\\\y = - [ \frac{1}{t + c} + t ][/tex]
Sixth Step: Apply the initial value problem and solve for the constant of integration ( c )
- We will use the given initial value statement i.e y ( 3 ) = 6 and evaluate the constant of integration ( c ) as follows:
[tex]y ( 3 ) = - [ \frac{1}{3 + c} + 3 ] = 6 \\\\\frac{1}{3 + c} = -9\\\\3 + c = -\frac{1}{9} \\\\c = - \frac{28}{9}[/tex]
Seventh Step: Express the solution of the ODE in an explicit form ( if possible ):
[tex]y = - [ \frac{1}{t - \frac{28}{9} } + t ][/tex]
