91Ó°ÊÓ

Let's define a new variable $$z=y^{\prime}$$, where z is a function of t. Taking the derivative of z with respect to t gives us $$z^{\prime}=y^{\prime \prime}$$. Now, the ODE can be written as two first-order ODEs: $$ y^{\prime}=z$$ $$z^{\prime}= t + z*z$$ with the initial conditions: $$y(1)=2$$ $$z(1)=1$$

To solve this system, we need to find the general solutions for both y(t) and z(t). Let's first solve for z(t). We can rewrite the second equation as: $$\frac{dz}{dt} - z^2 = t$$ This is a Bernoulli equation of the form: $$\frac{dz}{dt} + P(t)z = Q(t)z^n$$ where $$P(t)=0$$, $$Q(t)=t$$, and $$n=-1$$. Now, in order to solve the equation for z, we need to introduce a new variable v. We have: $$z=v^{-1}$$ $$\frac{d \left(v^{-1}\right)}{dt} = - v^{-2} \frac{dv}{dt}$$ By replacing these expressions into the Bernoulli equation, we get: $$- v^{-2} \frac{dv}{dt} = t$$ or equivalently: $$\frac{dv}{dt}=-tv^2$$ This is a separable equation: $$\frac{dv}{v^2}=-t dt$$ Integrate both sides of the equation, where a and b are constants of integration: $$ \int \frac{1}{v^2} dv = \int -t dt $$ $$ -a+\frac{-1}{v} = \frac{-t^2}{2} + b $$ Now, since $$z = v^{-1}$$, the equation becomes: $$a + z = \frac{t^2}{2} - b$$ Now, let's solve the first equation of the system, which is $$y^{\prime} = z$$. $$ \frac{dy}{dt} = a + z - \frac{t^2}{2} + b $$ Integrate both sides with respect to t: $$ y(t) = at + cz + \frac{- t^3}{6} + bt + C $$ where C is a constant of integration. Now we have the general solution for y(t).

Using the initial conditions, $$y(1)=2$$ and $$z(1)=1$$, we can find the constants a, b, c, and C. First, let's find the value of a: $$ 1 = a + 1 - \frac{1^2}{2} + b $$ $$ a = \frac{1}{2} - b $$ Now, we can use the initial condition $$y(1)=2$$: $$ 2 = a + c + \frac{-1^3}{6} + b $$ By replacing the a equation, we get: $$ 2 = \frac{1}{2} - b + c - \frac{1}{6} + b $$ $$ c= \frac{13}{6} $$ Now that we have the values of a, b, and c, we can obtain the specific solution for y(t): $$ y(t) = a(t) + c(z) + \frac{- t^3}{6} + b(t) + C $$ $$ y(t) = \left(\frac{1}{2} - b\right)t + \frac{13}{6}(1 - \frac{t^2}{2} + b) + \frac{- t^3}{6} + bt + C $$