Toronto Math Forum

\begin{equation}
-ysinx+y^3cosx+(3cosx+5y^2sinx)y'=0
\end{equation}
\begin{equation}
(-ysinx+y^3cosx)dx+(3cosx+5y^2sinx)dy=0
\end{equation}let M = -ysinx+y^3cosx, and N = 3cosx+5y^2sinx
\begin{equation}
M_y = -sinx+3y^2cosx, N_x = -3sinx+5y^2cosx
\end{equation} let My-Nx
\begin{equation}
M_y - N_x = 2sinx-2y^2cosx
\end{equation}since this looks famillier with M , so we are taking R1
\begin{equation}
R_1 = \frac{-M_y+N_x}{M}, R_1 = \frac{-2sinx+2y^2cosx}{-ysinx+y^3cosx},R_1 = \frac{2}{y}
\end{equation} the the integrating factor 𝝻 will be the e to the power of integral of R1
\begin{equation}
𝝻 = e^{∫R_1dy},𝝻 = y^2
\end{equation}then we times y^2 in this equation ,we get
\begin{equation}
(-y^3sinx+y^5cosx)dx+(3y^2cosx+5y^4sinx)dy=0
\end{equation}therefore
\begin{equation}
𝛗 = ∫-y^3sinx+y^5cosxdx=y^3cosx+y^5sinx+h(y)
\end{equation} take the derivative on 𝛗 of y,we get
\begin{equation}
𝛗_y = 3y^2cosx+5y^4sinx+h'(y)=3y^2cosx+5y^4sinx
\end{equation}we will have h'(y) = 0, so h(y)=C is a constant
\begin{equation}
𝛗=y^3cosx+y^5sinx+C
\end{equation}
\begin{equation}
y^3cosx+y^5sinx+C=0
\end{equation}since we have
\begin{equation}
y(\frac{\pi_1}{4})=√2
\end{equation}solve for C
\begin{equation}
C =6
\end{equation}
\begin{equation}
y^3cosx+y^5sinx+6=0
\end{equation}

\begin{equation}
xy''-(2x+1)y'+(x+1)y=0
\end{equation}
\begin{equation}
y''-\frac{2x+1}{x}y'+\frac{x+1}{x}y=0
\end{equation}p(x)=(2x+1)/x
\begin{equation}
𝝻=e^{-∫-\frac{2x+1}{x}dx}
\end{equation}W=C𝝻
\begin{equation}
𝝻=Ce^{2x}x
\end{equation}W=C𝝻
\begin{equation}
W = Ce^{2x}x
\end{equation}let C =1,then we have
\begin{equation}
W{(y_1,y_2)}=xe^{2x}
\end{equation}for part b), to check that y1 is a solution then
\begin{equation}
y_1' = e^x,y_1'' = e^x
\end{equation}
\begin{equation}
xe^x-(2x+1)e^x+(x+1)e^x=0
\end{equation}so y1 is indeed a solution.
\begin{equation}
W_{(y_1,y_2)}=xe^{2x}=e^xy_2'-e^xy_2
\end{equation}
\begin{equation}
y_2'-y_2=xe^x
\end{equation}the integrating factor is
\begin{equation}
𝝻 = e^{-x}
\end{equation} multiply 𝝻 in this equation
\begin{equation}
e^{-x}y_2'-e^{-x}y_2=x
\end{equation}
\begin{equation}
(e^{-x}y_2)'=x
\end{equation}take integral on both side
\begin{equation}
e^{-x}y_2=\frac{1}{2}x^2
\end{equation}move all the x to one side
\begin{equation}
y_2 = \frac{1}{2}x^2e^x
\end{equation}part c) , the general solution is
\begin{equation}
W = C_1y_1+C_2y_2
\end{equation}
\begin{equation}
y = C_1e^x+\frac{C_2}{2}x^2e^x
\end{equation}
\begin{equation}
y' = C_1e^x+\frac{C_2}{2}(2xe^x+x^2e^x)
\end{equation}we have y(1) = 0 and y(1)'=e
\begin{equation}
C_1e+\frac{C_2}{2}e=0, C_1e+\frac{C_2}{2}(2e+e)=e
\end{equation}solve for C1 and C2,then we have
\begin{equation}
C_1 =-\frac{1}{2},C_2 = 1
\end{equation}
\begin{equation}
y=-\frac{1}{2}e^x+\frac{1}{2}x^2e^x
\end{equation}

find the general solution to the given equation
\begin{equation}
9y''+6y'+y=0
\end{equation}we get
\begin{equation}
9r^2+6r+1=0
\end{equation}solve for r, we have
\begin{equation}
3r+1=0, r = -1/3
\end{equation} so we have
\begin{equation}
y(t) = C1e^{-1/3}+C2te^{-1/3}
\end{equation}

\begin{equation}
(2xy^2+2y)+(2x^2y+2x)y'=0
\end{equation} write y' into dy/dx, then we get
\begin{equation}
(2xy^2+2y)dx+(2x^2y+2x)dy=0
\end{equation} Let M = 2xy^2+2y , N = 2x^2y+2x , we get
\begin{equation}
\frac{dM}{dy}= 4xy + 2 , \frac{dN}{dx} = 4xy + 2
\end{equation} Since dM/dy = dN/dx, therefore they are exact. So ∃ 𝛗(x,y) = ∫ M dx
\begin{equation}
𝛗(x,y) = ∫ M dx = ∫ 2xy^2 + 2y dx = x^2y^2 + 2xy + h(y)
\end{equation}d𝛗(x,y)/dy = N
\begin{equation}
4x^2y + 2x +h'(y) = 2x^2y + 2x
\end{equation} we get h'(y) = 0 => h(y) = C, C is a constant.At the end we got the answer
\begin{equation}
𝛗(x,y) = x^2y^2 + 2xy +C
\end{equation}