# Doubt from differential equation

**Samarth_2020**#1

1 Like

I can't see your pic completely @Samarth_2020

Plz upload a complete pic of your question. Thanks

**Tanmay_1**#7

Pls ignore alpha =1 calculated in post 3...refer post 6 for correct value of alpha...Tanmay

**Shiva_1**#8

@Tanmay_1 it is ( 1+alpha)e^x+ alpha , in your answer you missed multiplying e^x with alpha. Refer my solution.

**Hari_Shankar**#10

y''=y' \Rightarrow y' =c_1e^x \Rightarrow y= c_1e^x+c_2

From given conditions c_1+c_2=1 and

\displaystyle c_1e^x=c_1e^x+c_2 + \int_0^1(c_1e^x+c_2) and hence

2c_2+c_1(e-1)=0

Solving the system yields c_1=\dfrac{2}{3-e} and c_2=\dfrac{1-e}{3-e}

3 Likes