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can you solve the initial value problem for:

dy/dx= xy-2x+y-2 , y(2)= -1

Aug 19, 2019

#1
+23861
+2

can you solve the initial value problem for:

$$\dfrac{dy}{dx}= xy-2x+y-2,\ y(2)= -1$$

$$\begin{array}{|rclcl|} \hline \mathbf{\dfrac{dy}{dx}} &=&\mathbf{ xy-2x+y-2 } \\\\ \dfrac{dy}{dx} &=& x(y-2)+ (y-2) \\\\ \dfrac{dy}{dx} &=& (y-2)(x+1) \\\\ \dfrac{dy}{(y-2)} &=& (x+1)dx \\\\ \large{ \int } \dfrac{1}{(y-2)}\ dy &=&\large{ \int } (x+1)\ dx \\\\ \mathbf{\ln(y-2) + c_1} &=& \mathbf{\dfrac{x^2}{2} +x + c_2 } \\ \hline && \ln(y-2) + c_1 &=& \dfrac{x^2}{2} +x + c_2 \quad | \quad y(2) = -1 \\ && \ln(-1-2) + c_1 &=& \dfrac{2^2}{2} +2 + c_2 \\ && \ln(-3) + c_1 &=& 4 + c_2 \quad | \quad \boxed{\mathbf{c_1 = -\ln(-3)}} \\ && \ln(-3) -\ln(-3) &=& 4 + c_2 \\ && 0 &=& 4 + c_2 \\ && \mathbf{c_2} &=& \mathbf{-4} \\ \hline \mathbf{\ln(y-2) -\ln(-3)} &=& \mathbf{\dfrac{x^2}{2} +x -4 } \\ \ln\left(\dfrac{y-2}{-3}\right) &=& \dfrac{x^2}{2} +x -4 \\ \dfrac{y-2}{-3} &=& e^\left(\dfrac{x^2}{2} +x -4 \right) \\ y-2 &=& -3e^\left(\dfrac{x^2}{2} +x -4 \right) \\ \mathbf{ y(x)} &=& \mathbf{2-3e^\left(\dfrac{x^2}{2} +x -4 \right)} \\ \hline \end{array}$$

Aug 19, 2019
#2
+1

THANK YOU!

Guest Aug 19, 2019