Question

The equation of a curve passing through the point \[\left( {0,{\text{ }}0} \right)\] and whose differential equation is \[y' = {e^x}sinx\].

Answer 1

Hint: First we know about the curve. Then we integrate the function which indicates the curve. We get an equation with integration constant. After that we calculate the value of integration constant. Then substitute the value of integration constant in the equation.

Complete step-by-step answer:
Given: differential equation \[y' = {e^x}sinx\]
We can write the above equation as \[\dfrac{{dy}}{{dx}} = {e^x}sinx\] …….…….(i)\[\left[ {\because y' = \dfrac{{dy}}{{dx}}} \right]\]
After arranging equation (i), we get
\[dy = {e^x}\sin xdx\]
\[\int {dy} = \int {{e^x}\sin xdx} \]
\[y = \int {{e^x}\sin xdx} \].............…(ii)
On equation (ii), we get two functions and . So we use integration by parts with an indefinite integral method. Because integral will continue to be repeated.
Calculation of Indefinite integral of \[{e^x}\]and\[\sin x\]:
Let \[I = \int {{e^x}\sin xdx} \]
\[ \Rightarrow \int {{e^x}\sin xdx} = {e^x}\int {\sin xdx - } \int {\left[ {\dfrac{d}{{dx}}{e^x}\int {\sin xdx} } \right]} \]
\[ \Rightarrow I = - {e^x}\cos x + \int {{e^x}coxdx} \] …………..…..(iii)
Similarly for \[{e^x}\]and \[cox\], we use again indefinite integration.
Let \[I = \int {{e^x}\cos xdx} \]
\[ \Rightarrow I = {e^x}\sin x - \int {\left[ {{e^x}\sin x} \right]} dx\]
\[\int {{e^x}\cos xdx} = {e^x}\sin x - \int {\left[ {{e^x}\sin x} \right]} dx\]……………...(iv)
Substitute value of \[\int {{e^x}\cos xdx} \]from equation (iv) into equation (iii), we get
\[I = - {e^x}\cos x + {e^x}\sin x - I\]
\[ \Rightarrow 2I = - {e^x}\cos x + {e^x}\sin x\]
\[ \Rightarrow 2I = {e^x}\left( {\sin x - \cos x} \right)\]
\[\therefore I = \dfrac{{{e^x}\left( {\sin x - \cos x} \right)}}{2}\]
Hence, \[\int {{e^x}\sin xdx = } \dfrac{{{e^x}\left( {\sin x - \cos x} \right)}}{2} + C\]
\[y = \dfrac{{{e^x}\left( {\sin x - \cos x} \right)}}{2} + C\]………………..(v)
Here C is integration constant.
Given that the curve is passing through the point \[\left( {0,{\text{ }}0} \right)\]. Therefore, on substituting \[x = 0,y = 0\] in equation (v), we get
\[0 = \dfrac{{{e^0}\left( {\sin 0 - \cos 0} \right)}}{2} + C\]
\[ \Rightarrow 0 = - \dfrac{1}{2} + C\]
\[\therefore C = \dfrac{1}{2}\]
Substitute value of C in equation (v) , we get
\[y = \dfrac{{{e^x}\left( {\sin x - \cos x} \right)}}{2} + \dfrac{1}{2}\]
\[ \Rightarrow 2y = {e^x}\left( {\sin x - \cos x} \right) + 1\]
\[ \Rightarrow \left( {2y - 1} \right) = {e^x}\left( {\sin x - \cos x} \right)\]

Hence, Equation of curve is \[ \left( {2y - 1} \right) = {e^x}\left( {\sin x - \cos x} \right)\].

Note: If \[f(x)\] and \[g(x)\] are two functions then integration of multiplication of two functions is obtained by integration by parts to solve integration.
\[\int {f(x)g(x)dx} = f(x)\int {g(x)dx - \int {\left[ {f'(x)\int {g(x)dx} } \right]dx} } \]
Indefinite integral: This type of integration is also known as antidifferentiation. It is called antidifferentiation because this process is the reversing of the process of differentiation.